The answer is 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.
A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.
Write a balanced equation.
Calculate the moles of H₂ in 30.8 g.
Calculate the moles of N₂ required to react with H₂.
Calculate the mass of N₂.
Calculate the initial mass of N₂.
Start with a balanced equation.
N₂ + 3H₂ --> 2NH₃
Calculate the moles of H₂ in 30.8 g.
n = m/M; where n = moles, m = mass, and M = molar mass.
M(H₂) = 1.008 g/mol
n(H₂) = (30.8 g)/(1.008 g/mol) = 30.56 mol H₂
Calculate the moles of N₂ required to react with 30.56 mol H₂ , using the mole ratio between H₂ and N₂ in the balanced equation.
30.56 mol H₂ × 1 mol N₂/3 mol H₂ = 10.18 mol N₂
Calculate the mass of N₂ in 10.18 mol.
m = n × M
M(N₂) = 2 × 14.007 g/mol N = 28.014 g/mol N₂
m(N₂) = 10.18 mol × 28.014 g/mol = 285.33g N₂
Therefore 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.
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