Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, accord

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 B

How many grams of NH3 can be produced from 3.89 mol of N2 and excess H2.

Express your answer numerically in grams.

Part C

How many grams of H2 are needed to produce 10.15 g of NH3?

Express your answer numerically in grams.

Part D

How many molecules (not moles) of NH3 are produced from 3.43×104 g of H2?

Express your answer numerically as the number of molecules.

Explanation / Answer

B)
1 mol of N2 produces 2 moles of NH3
so,
moles of NH3 = 2*moles of N2
= 2* 3.89 mol
= 7.78 mol

mass of NH3 = number of mol * molar mass
= 7.78 mol * 17 g/mol
= 132 g
Answer: 132 g

C)
moles of NH3 = mass/molar mass
= 10.15/17
= 0.597 mol

3 mol of H2 produces 2 moles of NH3
so,
moles of NH3 = (2/3)*moles of H2
0.597 = (2/3)*moles of H2
moles of H2 = 0.896 mol
mass of H2 = moles * molar mass
= 0.896 mol * 2 g/mol
= 1.79 g
Answer: 1.79 g

D)
mass of H2 = mass/molar mass
= (3.43*10^-4)/(2)
=1.715*10^-4 mol

3 mol of H2 produces 2 moles of NH3
so,
moles of NH3 = (2/3)*moles of H2
= (2/3)*1.715*10^-4 mol
=1.143*10^-4 mol

number of molecules of NH3 = moles * Avogadro's number
= 1.143*10^-4 * 6.022*10^23
= 6.88*10^19 molecules
Answer: 6.88*10^19 molecules

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