Question

For the balanced equation 2Al(s) + 3CuCl2(aq) -> 2Cu(s) + 3AlCl3(aq)
How many grams of CuCl2 are needed to react with 48.5 g of Al?

Answer 1

Answer:

361gm

Explanation:

gram molecular weight of Al = 27gm

gm mol. weight of CuCl2 = 134gm

From the reaction

2 ×27 gm of Al reacts with 3 × 134 gm of CuCl2

if so, then 48.5gm of Al reacts with  [(3 × 134) × (48.5) ] ÷ (2 ×27) gm

i.e., 361 gm of CuCl2