How many grams of N2 are required to produce 100.0 L of NH3 at STP?

Given:

N2, volume of 1L of NH3 at STP

Required:

Grams of N2

Solution:

N2 + 3H2 - > 2NH3

Molar mass of N2 = 28g

From the ideal gas equation PV = nRT

n = PV/RT

n = (1 atm) (100 L) / (0.08206 L-atm/mol-K) (273K)

n = 4.46 mol of NH3

from the reaction, we need 2 moles of NH3 to get 1 mole of N2

4.46 mol NH3 (1 mol N2/2 mol NH3) = 2.23 moles N2

2.23 moles of N2 (28 g N2/1 mol N2) = 62.5g N2

There is no equation given so I'll assume it's just the basic equation: N2 + 3H2 - > 2NH3

Molar mass of N2 = 28g

Since the two data sets are not in the same unit we first need to find the moles of gas from the given volume so we use: PV = nRT

where: P = 1 atm, V = 100ltr, T = 273 K

n = (1 atm) (100 L) / (0.08206 L-atm/mol-K) (273K) n = 4.46 mol of NH3

from the reaction equation we can see that 1 mole of N2 yields 2 moles of NH3 therefore:

4.46 mol NH3 (1 mol N2/2 mol NH3) = 2.23 moles N2

2.23 moles of N2 (28 g N2/1 mol N2) = 62.5g N2