Using English to Predict rendement of product a reaction

Predicting Reaction Products

When cooking, it's frequently handy to predict what will happen when we mix a bunch of ingredients together. For example, if we're interested in making a delicious new salad dressing, we would have a very small chance of making anything edible if we had no way of knowing which ingredients would have the greatest chance of succeeding.

Likewise, it's often necessary for chemists to predict the chemical reactions that will take place when two chemicals are combined. For example, if we're adding a chemical to a tank of toxic waste to stabilize it, we'd be very unhappy if we failed to predict an explosive reaction.

The Mole Says

The tips in this section, while helping you to figure out what reaction might occur, aren't infallible in correctly predicting the reaction that will take place. However, if you're not sure what will happen, these tips will be useful in suggesting some possibilities.

An easy way to predict what reaction will take place when two chemicals are mixed is to identify the type of reaction that's likely to occur when the chemicals are combined. Of course, we mentioned before that these types of reaction are arbitrary, but they do sometimes have a useful purpose.

Bad Reactions

There are two common mistakes when predicting the products of a chemical reaction. The first is predicting the formation of a theoretically impossible product such as NaCO₃ or Ag₄Cl. The second is failing to balance the equation once the products have been accurately predicted

Here are some tips you may find handy in helping to predict the type of reaction that will occur if you know only the reactants. Keep in mind that not all combiations of chemicals will result in a chemical reaction—these tips are handy only for helping to predict what would happen should they happen to react.

You've Got Problems

Problem 4: Write balanced chemical equations for the reactions that might occur when the following reactants are combined:
a) NaOH + H₂ SO₄ ⇔ ?
b) NH₃ + I₂ ⇔ ?
c) C₃H₈O + O₂ ⇔ ?

• If two ionic compounds are combined, it's usually safe to predict that a double displacement reaction will occur.
• If the chemicals mixed are oxygen and something containing carbon, it's usually a combustion reaction.
• If we start with only one reactant, the reaction taking place is probably a decomposition reaction. To predict the products of such a reaction, see what happens if the chemical breaks into smaller, familiar products such as water, carbon dioxide, or any of the gaseous elements.
• When pure elements are combined, synthesis reactions are the frequent result.
• If a pure element combines with an ionic compound, a single displacement reaction may take place.
• If a compound containing the hydroxide ion is involved, check the other compound to see if it contains hydrogen. If it does, it may be an acid-base reaction.

Rules For Predicting Products Of Chemical Reactions

·         Here are a few important things to remember when predicting products:

·         The compounds form must be neutral ionic compounds (which means you’ll be paying attention to their charges)

·         You do NOT carry subscripts from the reactants to the products.

·         You always balance your equation LAST

  

How to Calculate Percent Yield in Chemistry

In chemistry, the theoretical yield is the maximum amount of product a chemical reaction could create. In reality, most reactions are not perfectly efficient. If you perform the experiment, you'll end up with a smaller amount, the actual yield. To express the efficiency of a reaction, you can calculate the percent yield using this formula: %yield = (actual yield/theoretical yield) x 100. A percent yield of 90% means the reaction was 90% efficient, and 10% of the materials were wasted (they failed to react, or their products were not captured).

Part One of Three:
Finding the Limiting Reactant

1. 
   1
   . Start with a balanced chemical equation. A chemical equation describes the reactants (on the left side) reacting to form products (on the right side). Some problems will give you this equation, while others ask you to write it out yourself. Since atoms are not created or destroyed during a chemical reaction, each element should have the same number of atoms on the left and right side.^[1]
   

   • For example, oxygen and glucose can react to form carbon dioxide and oxygen: 6O_{2}+C_{6}H_{{12}}O_{6} → 6CO_{2}+6H_{2}O
     Each side has exactly 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 18 oxygen (O) atoms. The equation is balanced.
2. 
   2
   Calculate the molar mass of each reactant. Look up the molar mass of each atom in the compound, then add them together to find the molar mass of that compound. Do this for a single molecule of the compound.
   

   • For example, one molecule of oxygen (O_{2}) contains two oxygen atoms.
   • Oxygn's molar mass is about 16 g/mol. (You can find a more precise value on a periodic table.)
   • 2 oxygen atoms x 16 g/mol per atom = 32 g/mol of O_{2}.
   • The other reactant, glucose (C_{6}H_{{12}}O_{6}) has a molar mass of (6 atoms C x 12 g C/mol) + (12 atoms H x 1 g H/mol) + (6 atoms O x 16 g O/mol) = 180 g/mol.
3. 
   3
   Convert the amount of each reactant from grams to moles. Now it's time to look at the specific experiment you are studying. Write down the amounts of each reactant in grams. Divide this value by that compound's molar mass to convert the amount to moles.^[2]
   

   • For example, say you started with 40 grams of oxygen and 25 grams of glucose.
   • 40 g O_{2} / (32 g/mol) = 1.25 moles of oxygen.
   • 25g C_{6}H_{{12}}O_{6} / (180 g/mol) = about 0.139 moles of glucose.
4. 
   4
   Find the ratio of your reactions. Remember, a mole is just a large number that chemists use to "count" molecules. You now know how many molecules of each reactant you started with. Divide the moles of one reactant with the moles of the other to find the ratio of the two molecules.
   

   • You started with 1.25 moles of oxygen and 0.139 moles of glucose. The ratio of oxygen to glucose molecules is 1.25 / 0.139 = 9.0. This means you started with 9 molecules of oxygen for every 1 molecule of glucose.
5. 
   5
   Find the ideal ratio for the reaction. Go back to the balanced equation you wrote down earlier. This balanced equation tells you the ideal ratio of molecules: if you use this ratio, both reactants will be used up at the same time.
   

   • The left side of the equation is 6O_{2}+C_{6}H_{{12}}O_{6}. The coefficients tell you there are 6 oxygen molecules and 1 glucose molecule. The ideal ratio for this reaction is 6 oxygen / 1 glucose = 6.0.
   • Make sure you list the reactants in the same order you did for the other ratio. If you use oxygen/glucose for one and glucose/oxygen for the other, your next result will be wrong.
6. 
   6
   Compare the ratios to find the limiting reactant. In a chemical reaction, one of the reactants gets used up before the others. This limiting reactant determines how long the chemical reaction can take place. Compare the two ratios you calculated to identify the limiting reactant:^[3]
   

   • If the actual ratio is greater than the ideal ratio, then you have more of the top reactant than you need. The bottom reactant in the ratio is the limiting reactant.
   • If the actual ratio is smaller than the ideal ratio, you don't have enough of the top reactant, so it is the limiting reactant.
   • In the example above, the actual ratio of oxygen/glucose (9.0) is greater than the ideal ratio (6.0). The bottom reactant, glucose, must be the limiting reactant.

Part Two of Three:
Calculating Theoretical Yield

1. 
   
   • Continuing the example above, you are analyzing the reaction 6O_{2}+C_{6}H_{{12}}O_{6} → 6CO_{2}+6H_{2}O. The right hand side lists two products, carbon dioxide and water. Let's calculate the yield of carbon dioxide, CO_{2}.
   • 1. Identify your desired product. The right side of a chemical equation lists the products created by the reaction. Each product has a theoretical yield, meaning the amount of product you would expect to get if the reaction is perfectly efficient.
2. 
   2
   Write down the number of moles of your limiting reactant. The theoretical yield of an experiment is the amount of product created in perfect conditions. To calculate this value, begin with the amount of limiting reactant in moles. (This process is described above in the instructions for finding the limiting reactant.)
   

   • In the example above, you discovered that glucose was the limiting reactant. You also calculated that you started with 0.139 moles of glucose.
3. 
   3
   . Find the ratio of molecules in your product and reactant. Return to the balanced equation. Divide the number of molecules of your desired product by the number of molecules of your limiting reactant.
   

   • Your balanced equation is 6O_{2}+C_{6}H_{{12}}O_{6} → 6CO_{2}+6H_{2}O. There are 6 molecules of your desired product, carbon dioxide (CO_{2}). There is 1 molecule of your limiting reactant, glucose (C_{6}H_{{12}}O_{6}).
   • The ratio of carbon dioxide to glucose is 6/1 = 6. In other words, this reaction can produce 6 molecules of carbon dioxide from one molecule of glucose.
4. 
   4.  
   Multiply the ratio by the reactant's quantity in moles. The answer is the theoretical yield of the desired product in moles.
   

   • You started with 0.139 moles of glucose and the ratio of carbon dioxide to glucose is 6. The theoretical yield of carbon dioxide is (0.139 moles glucose) x (6 moles carbon dioxide / mole glucose) = 0.834 moles carbon dioxide.
5. 
   5
   Convert the result to grams. Multiply your answer in moles by the molar mass of that compound to find the theoretical yield in grams. This is a more convenient unit to use in most experiments.
   

   • For example, the molar mass of CO₂ is about 44 g/mol. (Carbon's molar mass is ~12 g/mol and oxygen's is ~16 g/mol, so the total is 12 + 16 + 16 = 44.)
   • Multiply 0.834 moles CO₂ x 44 g/mol CO₂ = ~36.7 grams. The theoretical yield of the experiment is 36.7 grams of CO₂.

Part Three of Three:
Calculating Percent Yield

1. 
   1. 
   Understand percent yield. The theoretical yield you calculated assumes that everything went perfectly. In an actual experiment, this never happens: contaminants and other unpredictable problems mean that some of your reactants will fail to convert to the product. This is why chemists use three different concepts to refer to yield:
   

   • The theoretical yield is the maximum amount of product the experiment could make.
   • The actual yield is the actual amount you created, measured directly on a scale.
   • The percent yield = {\frac {ActualYield}{TheoreticalYield}}*100\%. A percent yield of 50%, for instance, means you ended up with 50% of the theoretical maximum.
2. 
   2
   . Write down the actual yield of the experiment. If you performed the experiment yourself, gather the purified product from your reaction and weigh it on a balance to calculate its mass. If you are working from a homework problem or someone else's notes, the actual yield should be listed.
   

   • Let’s say our actual reaction yields 29 grams of CO₂.
3. 
   3
   . Divide the actual yield by the theoretical yield. Make sure you use the same units for both values (typically grams). Your answer will be a unit-less ratio.
   

   • The actual yield was 29 grams, while the theoretical yield was 36.7 grams. {\frac {29g}{36.7g}}=0.79.
4. 
   4
   Multiply by 100 to convert to a percentage. The answer is the percent yield.
   

   • 0.79 x 100 = 79, so the percent yield of the experiment is 79%. You created 79% of the maximum possible amount of Co2. 
5. 
   

   • Most Common Student Mistakes When Predicting And Writing Reaction Products In General Chemistry

1. Writing H(OH) (aq) as a product of the reaction of an acid and base. Hopefully, you realized that this product is H₂O(l).
2. Writing the incorrect molecular formula for a product. For example aluminum nitrate is often written as AlNO₃, not correctly as Al(NO₃)₃. To write this correctly, you need to know either that
• NO₃ has a negative one charge (predictable if you know the formula for nitric acid - HNO₃ - and realize that nitrate, the negative ion produced when a base abstracts a proton from the neutral acid , must be NO₃^-,
• Al ion has a charge of 3+ in combined with a nonmetal in a salt.
3. Not writing the simplest formula for a product. Fox example, sodium nitrate is not Na₂(NO₃)₂, but rather NaNO₃.
4. Leaving off the (aq), (s), (l), or (g) after the product. How can you or I tell if you have a precipitation reaction if one of the products doesn't have an (s) after it?
5. Not balancing the equations.
6. Not recognizing that if you have an acid as one reactant and a salt of the form M_x(OH)_y  as another reactant, then the reaction is an acid/base reaction.
7. Automatically assuming that if you have a strong acid in the reactants that the reaction has to be an acid/base reaction. This is only true if you have a base like OH^- in the reactants.  Acids can also act as oxidizing agents (for example when they dissolve pure metals) or "precipitating agents" when they provide an anion which forms a precipitate with a metal cation.
8. Recognize the that when -OH is covalently linked to a C, a nonmetal, IT IS NOT A BASE.  In fact it might be an acid if it is part of an oxyacid like nitric acid (HNO₃), a strong acid, or acetic acid (CH₃COOH), a weak acid.  C-OH is found often in organic carbon compounds, when it is called an alcohol.  
9. Assuming a reaction is a redox reaction without determining the oxidation number for each atom. (Obviously this is not necessary when the reaction is a combustion reaction using molecular oxygen, O₂.
10. Not knowing the charge on ions and not putting () around certain ions when needed. For example iron hydroxide is Fe(OH)₃ not FeOH₃.