Question

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: N2 1 g 2 1 3H2 1 g 2 h 2NH3 1 g 2

a. What is the maximum mass of ammonia that can be produced from a mixture of 1.00 3 103 g N2 and 5.00 3 102 g H2

Answer 1

Answer:

Mass of NH3 produced = 1217 g or 1.217*10^3 g

Explanation:

Given:

Mass of N2 = 1.003*10^3 g

Mass of H2 = 5.003*10^2 g

To determine:

Maximum mass of NH3 that can be produced when the given amounts of N2 and H2 combine

Calculation:

The chemical reaction corresponding to the production of ammonia is:

$N2(g)+3H2(g)\rightarrow 2NH3(g)$

Based on the reaction stoichiometry:

1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3

$moles\ of\ N2 = \frac{Mass\ N2}{Molar\ mass N2} = \frac{1.003*10^{3}g }{28g/mol} =35.8\ moles$

$moles\ of\ H2 = \frac{Mass\ H2}{Molar\ mass H2} = \frac{5.003*10^{2}g }{2g/mol} =250\ moles$

Since the moles of N2 is less than that of H2, the limiting reagent will be N2 which would in turn determine the amount of NH3 formed.

Based on the reaction stoichiometry the N2 : NH3 ratio = 1:2

Therefore,

$moles\ of\ NH3\ produced = 2*35.8 = 71.6\ moles\\\\Mass\ of\ NH3\ produced = moles*molar mass = 71.6\ moles*17\ g/mol = 1217 g$