10 g of Cu, 10 g of O2 react. Which is the limiting reactant? What amount of reactant remains in excess? Calculate the number of particles of reactant in excess. What is the mass of the produced product?

1 Answer
May 14, 2018

Answer:

Cu is the limiting reactant
#5 g " "O_2# in excess
#~~15 g " "CuO_2# produced

Explanation:

Given: #10# g Cu reacts with #10# g #O_2#.

Skeletal equation: #" "Cu + O_2 -> CuO_2#

balanced equation is the same as the skeletal equation.

molar masses:

#63.546 g/(mol) " "Cu; " " 2(15.999) = 31.998 g/(mol) " "O_2#

#63.546 + 31.998 = 95.544 g/(mol) " "CuO_2#

Find the limiting reagent:

#10 g Cu xx (1 mol Cu)/(63.546 g Cu) xx (1 mol CuO_2)/(1 mol Cu) = .1574 mol " " CuO_2#

#10 g O_2 xx (1 mol O_2)/(31.998g O_2) xx (1 mol CuO_2)/(1 mol Cu) = .1325 mol " " CuO_2#

Cu is the limiting reagent.

#O_2# is in excess.

Find the amount of #O_2# used:
#.1574 mol " " CuO_2 xx (1 mol O_2)/(1 mol CuO_2) xx (31.998 g O_2)/(1 mol O_2) ~~ 5 g " "O_2 #

Find the mass of #O_2# in excess:
#10 g - 5 g ("used")= 5 g# in excess

Find the mass of produced product #CuO_2#:

#.1574 mol " " CuO_2 xx (95.544 g CuO_2)/(1 mol CuO_2) = 15.04 g " "CuO_2#