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How many grams of ammonia will be formed upon the reaction of 30.7 grams of nitrogen gas with excess hydrogen gas

How many grams of ammonia will be formed upon the reaction of 30.7 grams of nitrogen-example-1
User Tombatron
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Answer: 37.32

Step-by-step explanation:

Nitrogen gas (N2) - 28.02 g/mol
Hydrogen gas (H2) - 2.02 g/mol
Ammonia gas (NH3) - 17.03 g/mol
Using the given mass of nitrogen gas, we can first calculate the number of moles of nitrogen gas present:

moles of N2 = mass of N2 / molar mass of N2


moles of N2 = 30.7 g / 28.02 g/mol

moles of N2 = 1.096 mol

From the balanced chemical equation, we know that one mole of nitrogen gas reacts with three moles of hydrogen gas to form two moles of ammonia gas. Since we have an excess of hydrogen gas, we can assume that all of the nitrogen gas will react to form ammonia gas. Therefore, we can calculate the number of moles of ammonia gas formed using the mole ratio from the balanced chemical equation:

moles of NH3 = (moles of N2) x (2 moles of NH3 / 1 mole of N2)

moles of NH3 = 1.096 mol x (2 mol/1 mol)

moles of NH3 = 2.192 mol

Finally, we can calculate the mass of ammonia gas formed using its molar mass:

mass of NH3 = moles of NH3 x molar mass of NH3

mass of NH3 = 2.192 mol x 17.03 g/mol

mass of NH3 = 37.32 g

User Full Stack Alien
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