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 N2 are necessary to produce 7.50 g of NH3?

Question

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 [latex]N_{2}[/latex] are necessary to produce 7.50 g of [latex]NH_{3}[/latex] ?

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 [latex]NH_{3}[/latex] are in 7.50 g ofv[latex]NH_{3}[/latex] . To convert grams of [latex]NH_{3}[/latex] to moles of [latex]NH_{3}[/latex] , we use the conversion factor 17.0 g [latex]NH_{3}[/latex] = 1 mol [latex]NH_{3}[/latex] . We see from the balanced chemical equation that 2 mol [latex]NH_{3}[/latex] are produced from 1 mol of N2, which gives us the conversion factor 2 mol [latex]NH_{3} [/latex] = 1.0 mol [latex]N_{2}[/latex]. Finally we convert moles of [latex]N_{2}[/latex] to grams of [latex]N_{2}[/latex], using the conversion factor 1 mol N2 = 28.0 g N2. Thus solving this example requires three steps and three conversion factors.

Accepted Answer

Step 1: Convert 7.50 grams of [latex]NH_{3}[/latex] to moles of [latex]NH_{3}[/latex] .

[latex] 7.50 \cancel{g NH_{3} } imes \frac{1 mol NH_{3} }{17.0 \cancel{g NH_{3} }} = mol NH_{3}[/latex]

Step 2: Convert moles of [latex]NH_{3}[/latex] to moles of [latex]N_{2}[/latex]

[latex] 7.50 \cancel{g NH_{3} } imes \frac{1 \cancel{mol NH_{3}} }{17.0 \cancel{g NH_{3} }} imes \frac{1 mol N_{2} }{2\cancel{mol NH_{3}} } = mol N_{2}[/latex] .

Step 3: Convert moles [latex]N_{2}[/latex] to grams of [latex]N_{2}[/latex] and now do the math.

[latex]7.50 \cancel{g NH_{3} } imes \frac{1 \cancel{mol NH_{3}} }{17.0 \cancel{g NH_{3} }} imes \frac{1\cancel{mol N_{2}} }{2\cancel{mol NH_{3}} } imes \frac{28.0 g N_{2} }{ 1 \cancel{mol N_{2}}} = 6.18 g N_{2}[/latex]

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 have just used.

Question 4.10: Stoichiometry . Ammonia is produced on an industrial scale b...

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