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Find the percent sulfate in a 1.50-gram sample of K\(_2\)SO\(_4\) if 1.05 grams of BaSO\(_4\) (233.3 grams/mole) is obtained. Use \( K = 39.1 \, \text{grams/mole} \) and \( SO_4^{2-} = 96.0 \, \text{grams/mole} \).

Answer :

Final answer:

First, calculate the molar mass of Potassium Sulfate and Barium sulfate. This can be used to find the mole of Barium sulfate created. The percent of sulfate in the sample can be calculated by dividing the mass of Barium sulfate by the mass of Barium sulfate produced times 100%. The percentage found is 52.2%.

Explanation:

To find the percent Sulfate in a sample of K2SO4, we first need to calculate the molar mass of K2SO4 and BaSO4. The formula for sulfate (SO42-) is BaSO4. The molar mass of BaSO4 is 233.3 g/mole, and the molar mass of K2SO4 is the sum of the molar masses of its component atoms.

Since 78.2 g (2 * 39.1 g K) + 96 g (SO4) = 174.2 g/mole for K2SO4, 1.50 g of K2SO4 equals 1.50 g / 174.2 g/mole = 0.00861 moles of K2SO4.

It is equivalent to 0.00861 moles of sulfate (SO42-), as each molecule of K2SO4 contains one sulfate ion. Therefore, 0.00861 moles of BaSO4 weigh 0.00861 moles x 233.3 g/mole = 2.01g.

The percent Sulfate is then the mass of BaSO4 divided by the mass of BaSO4 produced times 100%, which is 1.05 g / 2.01 g * 100% = 52.2%

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