Question

If 13.3 moles of Cu and 43.2 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

`=7.7molHNO_3`

Explanation:

Write the balanced equation and the given data.

`(color(red)(3)Cu)/color(red)(13.3mol)+(color(blue)8HNO_3)/color(blue)(43.2mol)->3Cu(NO_3)_2+2NO+4H_2O`

Determine how many moles each reactant needed if the reaction goes into completion. Refer to the balanced equation for the mole ratios; i.e.,

`ul(etaCu=13.3mol)`

`=13.3cancel(molCu)xx(color(blue)(8)molHNO_3)/(color(red)(3)cancel(molCu))`

`=35.5molHNO_3`

`:.`

`color(red)(13.3molCu)-=color(blue)(35.5molHNO_3)`

`color(blue)((etaHNO_3 " available")/(43.2mol)>(etaHNO_3" required")/(35.5mol))`

`ul(etaHNO_3=43.2mol)`

`=43.2cancel(molHNO_3)xx(color(red)(3)molCu)/(color(blue)(8)cancel(molHNO_3))`

`=16.2molCu`

`:.`

`color(blue)(43.2molHNO_3)-=color(red)(16.2molCu)`

`color(red)((etaCu " available")/(13.3mol)<(etaCu" required")/(16.2mol)`

Therefore, the `mol (eta) x's " reactant"` is computed as follows:

`eta " x's reactant "=etaHNO_3" available."-etaHNO_3" req'd"`

`eta " x's reactant "=43.2molHNO_3-35.5molHNO_2`

`eta " x's reactant "=7.7molHNO_3`