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.

the answer is alpha particles hope i you

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.