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

[tex]m_{CuCl}^{by\ Cu}=3.23molCu*\frac{2molCuCl}{2molCu}*\frac{99.0gCuCl}{1molCuCl} =320gCuCl\\\\m_{CuCl}^{by\ Cl_2}=1.64molCl_2*\frac{2molCuCl}{1molCl_2}*\frac{99.0gCuCl}{1molCuCl} =325gCuCl[/tex]

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.

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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

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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