Question

Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation N2(g) + 3H2(g) → 2NH3(g) Calculate the mass of ammonia produced when 37.0 g of nitrogen react with 12.0 g of hydrogen. g NH3 Which is the excess reactant and how much of it will be left over when the reaction is complete? hydrogen nitrogen

Answer 1

Answer:

The excess reactant is $H_2$.

Left over = 1.9903 moles or 4.0122 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For $N_2$

Given mass = 37.0 g

Molar mass of $N_2$ = 28.0134 g/mol

Moles of $N_2$ = 37.0 g / 28.0134 g/mol = 1.3208 moles

Given: For $H_2$

Given mass = 12.0 g

Molar mass of $H_2$ = 2.0159 g/mol

Moles of $H_2$ = 12.0 g / 2.0159 g/mol = 5.9527 moles

According to the given reaction:

$N_2_{(g)}+3H_2_{(g)}\rightarrow 2NH_3_{(g)}$

1 mole of $N_2$ react with 3 moles of $H_2$

1.3208 moles of $N_2$ react with 3*1.3208 moles of $H_2$

Moles of $H_2$ reacted = 3.9624 moles

Available moles of $H_2$ = 5.9527 moles

Limiting reagent is the one which is present in small amount. Thus, $N_2$ is limiting reagent as $H_2$ is present in excess . (3.9624 < 5.9527)

The excess reactant is $H_2$.

Left over = 5.9527 - 3.9624 moles = 1.9903 moles

Mass = moles × Molar mass = 1.9903 moles × 2.0159 g/mol  = 4.0122 g

Left over is 4.0122 g of hydrogen.