When 0.200 grams of Al reacts with 15.00 mL of a 0.500 M copper (II) chloride solution, how many moles of solid Cu would be produced? How many grams of solid Cu would be produced? Show all calculations including the mole-to-mole ratios used in the calculation.

Question

Royce Sharp

Chemistry

6 January, 05:27

When 0.200 grams of Al reacts with 15.00 mL of a 0.500 M copper (II) chloride solution, how many moles of solid Cu would be produced? How many grams of solid Cu would be produced? Show all calculations including the mole-to-mole ratios used in the calculation.

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Answers (1)

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Kaiden Serrano · Answer 1

When The balanced equation is:

2Al + 3CuCl2 ⇒3 Cu + 2AlCl3

So, we want to find the limiting reactant:

1 - no. of moles of 2Al = MV/n = (Wt * V) / (M. Wt*n*V) = Wt / (M. Wt * n)

where M = molarity, V = volume per liter and n = number of moles in the balanced equation.

by substitute:

∴ no. of moles of 2Al = 0.2 / (26.98 * 2) = 0.003706 moles.

2 - no. of moles of 3CuCl2 = M*v / n = (0.5 * (15/1000)) / 3 = 0.0025 moles.

So, CuCl2 is determining the no. of moles of the products.

∴The no. of moles of 3Cu = 0.0025 moles.

∴The no. of moles of Cu = 3*0.0025 = 0.0075 moles.

and ∵ amount of weight (g) = no. of moles * M. Wt = 0.0075 * M. wt of Cu

= 0.0075 * 63.546 = 0.477 g