Question

2 Cu + Cl2 ----> 2 Cuci

If 1.64 moles of chlorine is reacted with 3.23 moles of copper, how many grams of copper I chloride will be made?
(The next question will ask about the limiting and excess reactants for this reaction)
o 320 g
O 160 g
O 162 g
O 325 g

Answer 1

320 g.

Step-by-step explanation:

Hello!

In this case, according to the balanced chemical reaction, we can compute the grams of copper I chloride produced by each reactant, as shown below:

`m_(CuCl)^(by\ Cu)=3.23molCu*(2molCuCl)/(2molCu)*(99.0gCuCl)/(1molCuCl) =320gCuCl\\\\m_(CuCl)^(by\ Cl_2)=1.64molCl_2*(2molCuCl)/(1molCl_2)*(99.0gCuCl)/(1molCuCl) =325gCuCl`

Thus, since copper produces the fewest grams of CuCl, we infer it is the limiting reactant, therefore the correct mass of copper I chloride is 320 g.

Best regards!