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12.07.2022 - 05:39

Determine the quantity (g) of pure MgSO4 in 3.2 g of MgSO4.7H2O.

Question:

Determine the quantity (g) of pure {eq}MgSO_4 {/eq} in 3.2 g of {eq}MgSO_4.7H_2O {/eq}.

Answers (1)

Jennie April 4, 2023 в 07:57

To determine the quantity of pure {eq}MgSO_4 {/eq} in 3.2 g of {eq}MgSO_4.7H_2O {/eq}, we need to first calculate the molar mass of {eq}MgSO_4.7H_2O {/eq}. Molar mass of {eq}MgSO_4.7H_2O {/eq} = (1 ? molar mass of Mg) + (1 ? molar mass of S) + (4 ? molar mass of O) + (7 ? molar mass of H2O) = (1 ? 24.31) + (1 x 32.06) + (4 ? 16.00) + (7 ? 18.02) = 246.47 g/mol Now, we can find the percentage of pure {eq}MgSO_4 {/eq} in {eq}MgSO_4.7H_2O {/eq} by dividing the molar mass of {eq}MgSO_4 {/eq} by the molar mass of {eq}MgSO_4.7H_2O {/eq} and multiplying by 100: Percentage of pure {eq}MgSO_4 {/eq} in {eq}MgSO_4.7H_2O {/eq} = (molar mass of {eq}MgSO_4 {/eq} / molar mass of {eq}MgSO_4.7H_2O {/eq}) ? 100 = (24.31 + 32.06 + 4 ? 16.00) / 246.47 ? 100 = 60.31% Therefore, the quantity of pure {eq}MgSO_4 {/eq} in 3.2 g of {eq}MgSO_4.7H_2O {/eq} can be calculated as: Quantity of {eq}MgSO_4 {/eq} = (3.2 g) ? (60.31 / 100) = 1.93 g Hence, there are 1.93 g of pure {eq}MgSO_4 {/eq} in 3.2 g of {eq}MgSO_4.7H_2O {/eq}.

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