What mass of salt can be obtained by the interaction of 200 g of a 20% solution of copper (II)

| September 9, 2021 | education

What mass of salt can be obtained by the interaction of 200 g of a 20% solution of copper (II) sulfate and 15 g of magnesium?

1. We make a chemical equation, not forgetting to put down the coefficients:

CuSO4 + Mg = MgSO4 + Cu.

2. Find the mass of CuSO4:

m (CuSO4) = m (solution) * ω = 200 * 0.2 = 40g.

3. Find the molar mass of CuSO4:

M (CuSO4) = 64 + 32 + 16 * 4 = 160 g / mol.

4. Find the amount of substance CuSO4:

n (CuSO4) = m / M = 40/160 = 0.25 mol.

5. Let’s find the amount of Mg substance:

n (Mg) = m / M = 15/24 = 0.625 mol.

6. We see that: n (Mg) <n (CuSO4); Therefore, further all calculations will be carried out for n (Mg).

7. Find the amount of substance CuSO4:

n (MgSO4) = n (CuSO4) = 0.25 mol.

8. Let’s find the molar mass of MgSO4:

M (MgSO4) = 24 + 32 + 16 * 4 = 120 g / mol.

9. Find the mass of MgSO4:

m (MgSO4) = n * M = 0.25 * 120 = 30g.

Answer: m (MgSO4) = 30g.



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