Question

For the following reaction, 5.21 grams of nitrogen gas are allowed to react with 5.91 grams of hydrogen gas.

Answer 1

Answer:

Maximum amount of ammonia that can be formed is 6.32 g

Explanation:

The balanced equation is:

N2 + 3H2\rightarrow 2NH3

$Moles\ N2 = \frac{5.21g}{28g/mol} =0.186\\\\Moles H2 = \frac{5.21g}{2g/mol} =2.61$

N2 is the limiting reactant

Based on the reaction stoichiometry:

1 mole N2 produced 2 moles NH3

therefore, 0.186 mol N2 will yield: 2(0.186) = 0.372 moles NH3

Molar mass NH3 = 17 g/mol

$Maximum\ Mass\ NH3 = 0.372 mols*17g/mol=6.32g$