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The methods described in the previous section allow us to express reactants and products in terms of moles, but what if we wanted to know how many grams of a reactant would be required to produce a given number of grams of a certain product? This logical extension is, of course, trivial! In Chapter 4, we learned to express molar quantities in terms of the masses of reactants or products. For example, the reduction of iron (III) oxide by hydrogen gas, produces metallic iron and water. If we were to ask how many grams of elemental iron will be formed by the reduction of 1.0 grams of iron (III) oxide, we would simply use the molar stoichiometry to determine the number of moles of iron that would be produced, and then convert moles into grams using the known molar mass. For example, one gram of Fe2O3 can be converted into mol Fe2O3 by remembering that moles of a substance is equivalent to grams of that substance divided by the molar mass of that substance: \[moles=\left ( \frac{grams}{molar\; mass} \right )=\left ( \frac{grams}{grams/mol} \right )=(grams)\times (mol/grams) \nonumber \] Using this approach, the mass of a reactant can be inserted into our reaction pathway as the ratio of mass-to-molar mass. This is shown here for the reduction of 1.0 gram of Fe2O3. \[Given:\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\; \; Find: x\: mol\: Fe \nonumber \] We set up the problem to solve for mol product; the general equation is: \[(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber \] The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting, \[x\: mol\: Fe=\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right ) \nonumber \] It is often simpler to express the ratio (mass)/(molar mass) as shown below, \[\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right ) \nonumber \] Doing this, and rearranging, \[x\: mol\: Fe=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.013\: mol \nonumber \] That is, the reduction of 1.0 grams of Fe2O3 by excess hydrogen gas will produce 0.013 moles of elemental iron. All of these calculations are good to two significant figures based on the mass of iron (III) oxide in the original problem (1.0 grams). Note that we have two conversion factors (ratios) in this solution; one from mass to molar mass and the second, the stoichiometric mole ratio from the balanced chemical equation. Knowing that we have 0.013 moles of Fe, we could now convert that into grams by knowing that one mole of Fe has a mass of 55.85 grams; the yield would be 0.70 grams. We could also modify our basic set-up so that we could find the number of grams of iron directly. Here we have simply substituted the quantity (moles molar mass) to get mass of iron that would be produced. Again, we set up the problem to solve for mol product; \[(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right ) \nonumber \] In place of mol product and mol reactant, we use the expressions for mass and molar mass, as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant (the given) will cancel, giving a solution in mol product. Substituting, \[x\: mol\: Fe\left ( \frac{55.85g\: Fe}{mol\: Fe} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right ) \nonumber \] Rearranging and canceling units, \[x\: g\: Fe=\left ( \frac{55.85g\: Fe}{mol\: Fe} \right ) \left ( \frac{1\: g\: Fe_{2}O_{3}\times mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.70g \nonumber \]
Aqueous solutions of silver nitrate and sodium chloride react in a double-replacement reaction to form a precipitate of silver chloride, according to the balanced equation shown below. AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq) If 3.06 grams of solid AgCl are recovered from the reaction mixture, what mass of AgNO3 was present in the reactants?
Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in below. 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) If 17.467 grams of chlorine gas are allowed to react with excess Al, what mass of solid aluminum chloride will be formed?
Ammonia, NH3, is also used in cleaning solutions around the house and is produced from nitrogen and hydrogen according to the equation: N2 + 3 H2 → 2 NH3
Miriel N. Consider the following reaction. 4Fe + 3O2 2Fe2O3 How many moles of O2 are required to react completely with 54.3 mg of Fe? How many grams Fe2O3, are produced from the complete reaction of 32.2 g of Fe? 2 Answers By Expert Tutors Britt P. answered • 03/13/17 College Professor for Chemistry and Math Tutoring.
The first question involves a mass of Fe, you will need the molar mass of Fe. [1 mol Fe=55.85 g Fe] The question asks about the moles of O2 but starts with information about Fe, you will need the mole ratio from the balanced chemical equation. [3 mol O2 = 4 mol Fe] Because molar mass is grams per mole and the question starts you with mg, you need the conversion between g and mg. ? moles O2 = 54.3 mg Fe x (10-3 g / 1 mg) x (1 mol Fe/55.85 g) x (3 mol O2 / 4 mol Fe) (54.3 * 10-3 *1*3)/(1*55.85*4) = 0.1629 / 223.4 = 0.000729 mol O2 How many grams Fe2O3, are produced from the complete reaction of 32.2 g of Fe? Question asks about grams of Fe2O3; you need the molar mass of iron(III) oxide. [1 mol Fe2O3 = 159.70 g] Question starts with 32.2 g Fe; you need the molar mass of Fe. [1 mol Fe = 55.85 g] Question involves both iron and iron(III) oxide; you need the mole ratio from the balanced chemical equation. [4 mol Fe = 2 mol Fe2O3] ?g Fe2O3 = 32.2 g Fe * (1 mol Fe/55.85 g Fe) * (2 mol Fe2O3 / 4 mol Fe)* (159.70 g / 1 mol Fe2O3) (32.2*1*2*159.7) / (55.85*4*1) = (10248.68) / (223.4) = 46.0 g Fe2O3
Hank L. answered • 03/13/17 Engaging Ivy League Grad / Teacher for Chemistry and Physics Tutoring!
This problem is only slightly more complicated than your other problem (KClO3) The balanced reaction is 4Fe + 3O2 = 2Fe2O3 The reason this is a little more complex is that the mole ratios aren't 1:1 like they were in the other problem. Here are the steps you should take to solve problems like this: 1. Balance the equation (you did that already) 2. Convert what you are given in grams into moles. In this case you'll want to convert 54.3 mg of Fe into moles 3. Find the mole ratio of what you're trying to find (in this case O2) divided by what you are given (in this case Fe). The mole ratio is simply a fraction using the coefficients from the balanced chemical reaction. In this case, you want O2 and are given Fe, thus your mole ratio is 3 O2 / 4 Fe 4. Multiply your answer from 2 by your mole ratio in 3 to get moles of what you're trying to find 5. Now you solved the number of moles of what you're trying to find, so convert it back to grams! Step 2: First convert mg to g: 54.3mg Fe = 0.0543g Fe Now, 0.0543g (1 mol Fe / 55.8g) = 9.73 x 10-4 moles of Fe Step 3: We found the mole ratio is 3O2/4Fe Step 4: multiply steps 2 and 3 together: 9.73 x 10-4mol of Fe * 3O2/4Fe = 7.30 x 10-4 mol O2 needed We don't need to do step 5 because the question wants the number of moles of O2 which is 7.30 x 10-4 To solve the second part, you start with 32.2g of Fe Step 2: Convert to moles: 32.2g Fe (1 mol Fe / 55.8g) = 0.577 mol Fe Step 3: Mole ratio of Fe2O3 to Fe is (2 Fe2O3 / 4 Fe) So the mole ratio is 2/4 Step 4: multiply steps 2 and 3: 0.577 * 2/4 = 0.289 mol Fe2O3 is formed Step 5: convert product back to grams: 0.289 mol Fe2O3 * ((2*55.8+3*16)grams / 1 mol Fe2O3) 0.289 mol Fe2O3 * ((2*55.8+3*16)grams / 1 mol Fe2O3) =0.289 * (159.6g) = 46.1 grams of Fe2O3 is produced |