If 25.0 g of carbon monoxide react with 8.50 g of ammonia and 10.0 g of hydrogen to produce water and acetonitrile (CH3CN), what mass of each excess reactant is left over after the reaction is complete?

The mass of NH₃ left over = 0.91 g of NH₃

The mass of H₂ left over = 8.2 g of H₂

Explanation:

The given information are;

The mass of the carbon monoxide present in the reaction = 25.0 g

The mass of the ammonia present in the reaction = 8.50 g

The mass of the hydrogen present in the reaction = 10.0 g

The above masses reacts to produce water and acetonitrile (CH₃CN)

The balanced chemical equation for the reaction is given as follows;

2CO + NH₃ + 2H₂ → CH₃CN + 2H₂O

Therefore, two moles of CO reacts with one mole of NH₃ and  two moles of H₂ to produce one mole of CH₃CN and two moles of H₂O

The molar mass of CO = 28.01 g/mol

The molar mass of NH₃ = 17.031 g/mol

The molar mass of H₂ = 2.0159 g/mol

The molar mass of H₂O = 18.015 g/mol

The molar mass of CH₃CN = 41.05 g/mol

The number of moles of CO present = 25/28.01 = 0.893 moles

The number of moles of NH₃ present = 8.5/17.031 = 0.5 moles

The number of moles of H₂ present = 10/2.0159 = 4.96 moles

Therefore, 0.893 moles of CO reacts with 0.893/2 mole of NH₃ and  0.893 moles of H₂ to produce 0.893/2 mole of CH₃CN and 0.893 moles of H₂O

The excess reactants left are;

0.5 - 0.893/2 = 0.0535 moles of NH₃ with mass 0.0535 × 17.031 = 0.91 g

4.96 - 0.893 = 4.067 moles of H₂ with mass 4.067 × 2.0159 = 8.2 g

The mass of NH₃ left over = 0.91 g of NH₃

The mass of H₂ left over = 8.2 g of H₂.

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