Answer:
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.
