Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, accord
ID: 900960 • Letter: H
Question
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation
3H2(g)+N2(g)2NH3(g)
Part A
How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2?
Express your answer numerically in moles.
Part B
How many grams of NH3 can be produced from 2.88 mol of N2 and excess H2.
Express your answer numerically in grams.
Part C
How many grams of H2 are needed to produce 13.26 g of NH3?
Part D
How many molecules (not moles) of NH3 are produced from 1.22×104 g of H2?
Explanation / Answer
Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation
3H2(g)+N2(g)2NH3(g)
Part A
How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2?
Express your answer numerically in moles.
if H2 is limiting reaction, the
ratio of H2 and NH3 is given by
3 mol of H2 = 2 mol of NH3
18 mol of H2 = x mol fo NH3
x = 18*2/3 = 12 mol of NH3
moles of NH3 possible = 12 mol
Part B
How many grams of NH3 can be produced from 2.88 mol of N2 and excess H2.
Express your answer numerically in grams.
1 mol of N2 = 2 mol of NH3
2.88 mol of N2 = x mo of NH3
x = 2*2.88 = 5.76 mol of NH3
1 mol fo NH3 = 17 g/mol
5.76 mol --> 17*5.76 = 97.92 g of NH3 can be produced
Part C
How many grams of H2 are needed to produce 13.26 g of NH3?
3 mol of H2 = 2 mol of NH3
6 g of H2 = 34 g of NH3
x g of H2 = 13.26 g of NH3
x = 13.26/34*6
x = 2.34 g of H2 will be required
Part D
How many molecules (not moles) of NH3 are produced from 1.22×104 g of H2?
6 g of H2 = 34 g of NH3
1.22*10^-4 g of H2 = x g of NH3
x = (1.22*10^-4)/6*34 = 0.0006913 g of NH3
mol of NH3 = mass/MW = 0.0006913/17 = 0.000040664 mol
1 mol = 6.022*10^23 molecules
molecules = 0.000040664*6.022*10^23 = 2.44878*10^22 molecules
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