Balanced equation: N2 + 3H2 -> 2NH3
Molar mass of NH3: 14 + 3(1) = 17 g/mol
For 14g of H2:
- Calculate moles of H2:
14g H2 * (1 mol H2 / 2g H2) = 7 moles H2
- Using the mole ratio from the balanced equation, calculate moles of NH3 produced:
7 moles H2 * (2 moles NH3 / 3 moles H2) = 4.67 moles NH3
- Calculate grams of NH3 produced:
4.67 moles NH3 * 17 g/mol = 79.39g
Therefore, if 14g of H2 are reacted with an excess of N2, 79.39g of NH3 will be produced.
Use the following reaction. Remember to first check if it is balanced!
N2+3H22NH3
nitrogen + hydrogen ammonia
Complete the stoichiometry table below to calculate how many grams of ammonia are produced if 14g of H₂ are reacted with an excess of N₂. You will need to calculate molar masses with the following information:
Substance
H
N
Molar mass
1.0 g/mol
14.0 g/mol
14 g H₂ 2
1 mol H2
g NH3
1.0 g H₂ 2
1 mol NH3
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