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35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--> 2NH3
a) Identify the limiting reagent:
b) Calculate the grams of ammonia formed:
c) Calculate the grams of excess reactant formed:

Respuesta :

ChoiSungHyun

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

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