Question

If 12.4 moles of Cu and 48.2 moles of HNO3 are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? 3 Cu + 8 HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Answer 1

Excess reactant amount 15.1 moles

Step-by-step explanation:

3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO +4 H₂O

Using mole ratio method we find limiting reagent and excess reagent

For,

`Cu(mole)/(stoichiometry) (12.4)/(3)=4.13`

`HNO_3{(mole)/(stoichiometry) } (48.2)/(8)=6.025`

Since, Cu is limiting reagent and HNO₃ excess reagent

According to reaction,

3 moles of Cu react with 8 moles of HNO₃

12.4 moles of Cu react with = 33.06 moles of HNO₃

So remaining amount of excess HNO₃ = 48.2 - 33.06

= 15.1 moles