The mass of copper released during the interaction of copper sulfate 2 with iron weighing 28 g?

univerkov | November 29, 2020 | education

Let’s write down given
CuSO4
m (Fe) = 28 g
Let’s write a reaction
CuSO4 + Fe = FeSO4 + Cu
Find the amount of iron
n = m / M
m – mass of matter
M- molar mass
n- amount of substance
n (Fe) = 22: 56 = 0.39 mol
n (Fe) = n (Cu)
m (Cu) = M • n = 64 • 0.39 = 24.96 g
Answer: 24.96 g of substance

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