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

1 Answer
Jarni Renz
Feb 3, 2018

#=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#

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