Question 4.12: Stoichiometry Ammonia is produced on an industrial scale by ...
Stoichiometry
Ammonia is produced on an industrial scale by the reaction of nitrogen gas with hydrogen gas (the Haber process) according to this balanced equation:
How many grams of \text{N}_{2} are necessary to produce 7.50 g of \text{NH}_{3}?
Strategy
The coefficients in an equation refer to the relative numbers of moles, not grams. Therefore, we must first find out how many moles of \text{NH}_{3} are in 7.50 g of \text{NH}_{3}. To convert grams of \text{NH}_{3} to moles of \text{NH}_{3}, we use the conversion factor 17.0 g \text{NH}_{3} = 1 mol \text{NH}_{3}. We see from the balanced chemical equation that 2 mol \text{NH}_{3} are produced from 1 mol \text{N}_{2}, which gives us the conversion factor 2 mol \text{NH}_{3} = 1.0 mol \text{N}_{2}. Finally, we convert moles of \text{N}_{2} to grams of \text{N}_{2}, using the conversion factor 1 mol \text{N}_{2} = 28.0 g \text{N}_{2}. Thus, solving this example requires three steps and three conversion factors.
Step 1: Convert 7.50 grams of \text{NH}_{3} to moles of \text{NH}_{3}.
7.50\ \text{g NH}_{3}\times \frac{1\ \text{mol NH}_{3}}{17.0\ \text{g NH}_{3}}= 0.441 \ \text{mol NH}_{3}Step 2: Convert moles of \text{NH}_{3} to moles of \text{N}_{2}.
0.441\ \text{mol NH}_{3}\times \frac{1\ \text{mol N}_{2}}{2\ \text{mol NH}_{3}}= 0.221 \ \text{mol N}_{2}Step 3: Convert moles \text{N}_{2} to grams of \text{N}_{2}.
0.221\ \text{mol N}_{2}\times \frac{28.0\ \text{g N}_{2}}{1\ \text{mol N}_{2}}= 6.18 \ \text{g N}_{2}Alternatively, one could perform the calculations in one continuous step.
7.50\ {g \cancel{NH}}_{3}\times \frac{1\ \cancel{\text{mol NH}_{3}}}{17.0\ \cancel{\text{g NH}_{3}}} \times \frac{1\ \cancel{\text{mol N}_{2}}}{2\ \cancel{\text{mol NH}_{3}}} \times \frac{28.0\ \text{g N}_{2}}{1\ \text{mol N}_{2}} = 6.18\ \text{g N}_{2}In all such problems, we are given a mass (or number of moles) of one compound and asked to find the mass (or number of moles) of another compound. The two compounds can be on the same side of the equation or on opposite sides. We can do all such problems by the three steps we just used. The remaining problems in this chapter involving multiple steps will be solved using the one continuous step method shown directly above.
■ Quick Check 4.12
Pure aluminum is prepared by the electrolysis of aluminum oxide according to this equation:
(a) Balance this equation.
(b) What mass of aluminum oxide is required to prepare 27 g (1 mol) of aluminum?