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Metallic copper is formed when aluminum reacts with copper(II) sulfate. How many grams of metallic copper can be obtained when 54.0 g of Al react with 319 g of CuSO4? Al + 3CuSO4→ Al2(SO4)3 +3Cu
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If calculate the amount of moles that are in 54 grams of AL and 319 grams of CuSO4, you'd notice that AL=27 grams, CuSo4=162 grams that's one half of what they want to react with. so you'd multiply the entire equation by 2. 3CU x 2=?
it should be 127.6 g Cu
CuSO4 required: 54 g Al * 1mol Al/27gr Al*3 mol CuSO4 /1 mol Al *160gr CuSO4/1 mol CuSO4 = 960 gr but we have 319 gr so CuSO4 is limiting then; m(CU) = 319 gr CuSO4*1molCuSO4/160 CuSO4*3mol cu/3mol CuSO4*64 g Cu/1 mol Cu = 127.6 gr
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