When 9.9 g of zinc hydroxide interacted with sulfuric acid, 12.88 g of salt was formed.

| July 28, 2021 | education

When 9.9 g of zinc hydroxide interacted with sulfuric acid, 12.88 g of salt was formed. Find the yield of the reaction product.

1. Let’s compose the equation for the reaction of zinc hydroxide with sulfuric acid:

Zn (OH) 2 + H2SO4 = ZnSO4 + 2H2O;

2. Calculate the chemical amount of zinc hydroxide:

n (Zn (OH) 2) = m (Zn (OH) 2): M (Zn (OH) 2);

M (Zn (OH) 2) = 65 + 2 * 17 = 99 g / mol;

n (Zn (OH) 2) = 9.9: 99 = 0.1 mol;

3. Determine the theoretical amount of zinc sulfate:

ntheor (ZnSO4) = n (Zn (OH) 2) = 0.1 mol;

4. Let’s calculate the theoretical mass of salt:

mtheor (ZnSO4) = ntheor (ZnSO4) * M (ZnSO4);

M (ZnSO4) = 65 + 32 + 4 * 16 = 161 g / mol;

mtheor (ZnSO4) = 0.1 * 161 = 16.1 g;

5. Let’s find the output of the product:

ν = mp (ZnSO4): mtheor (ZnSO4) = 12.88: 16.1 = 0.8 or 80%.

Answer: 80%.



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