Examine the equation below then answer the questions that follow it. N2(g) + 3H2(g) ==> 2NH3(g)

A. If 16.12 g of hydrogen is reacted at STP, volume of ammonia is produced?
B. If 24.15 L of nitrogen is reacted at STP, what mass of hydrogen must also be reacted?
C. At STP, what volume of hydrogen is needed to produce 46.9 L of ammonia
D. What volume of nitrogen is needed to produce 0.847 mol of ammonia?

A.

The molar mass of H2 ≈ 2 g/mol. First we need to find how many moles of H2 we have. (All of the calculations are essentially unit conversions).

16.12 g H2 * (1 mol H2 / 2 g H2 ) = 8.06 moles H2

Looking at the reaction we see that 3 moles of H2 produces 2 moles of NH3.

(2 mol NH3 / 3 mol H2) * (8.06 mol H2) = 5.37 mol NH3.

Now we can use the ideal gas equation to see what volume of NH3 we have at STP with 5.37 moles.

PV=nRT

STP => T=273 K, P = 1 atm (and R = 0.08205 L*atm / mole *K)

V = (nRT) / P = (5.37 * R * 273) / (1) = 120.3 L


B.

First find how many moles of N2 we have from the ideal gas equation.

PV = nRT

n = PV / RT = (1 * 24.15) / (R * 273) = 1.1 moles N2

From the reaction we 1 mole of N2 is needed to react with 3 moles of H2.

1.1 mol N2 * (3 mol H2 / 1 mol N2) = 3.3 mol H2

Molar mass of H2 = 2 g/mol

2 g/mol * (3.3 mol H2) = 6.6 g H2


C.

Find the number of moles of ammonia

n = PV / RT = (1 * 46.9) / (R * 273) = 2.1 moles NH3

3 moles of H2 produce 2 moles of NH3

(3 mol H2 / 2 mol NH3) * (2.1 mol NH3) = 3.2 mol H2

V = nRT / P = (3.2 * R * 273) / (1) = 71.7 L H2


D.

1 mole of N2 needed to produce 2 moles of NH3

0.847 mol NH3 * (1 mol N2 / 2 mol NH3) = 0.424 mols N2

V = nRT / P = (0.424 * R * 273) / (1) = 9.5 L

B*

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