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?

Md Ariful Islam

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Md Ariful Islam

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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 :

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

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

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 $CuO_2$:

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

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