A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified (NH₄)₂MoO₄ to produce a precipitate of (NH₄)₃PO₄·12MoO₃ (MM = 1877 g mol⁻¹). If 24.21 g of dry (NH₄)₃PO₄·12MoO₃ was obtained, what was the mass of sodium phosphate in the original sample?

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A sample was contaminated with sodium phosphate - HSC - SSCE Chemistry - Question 17 - 2021 - Paper 1

Question 17

A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified (NH₄)₂MoO₄ to produce a precipitate of (NH₄)₃P... show full transcript

Worked Solution & Example Answer:A sample was contaminated with sodium phosphate - HSC - SSCE Chemistry - Question 17 - 2021 - Paper 1

Step 1

Calculate the moles of (NH₄)₃PO₄·12MoO₃

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Answer

To find the moles of the precipitate, we use the formula:

ext{Moles} = rac{ ext{Mass}}{ ext{Molar Mass}}

Using the mass of the precipitate:

ext{Moles} = rac{24.21 ext{ g}}{1877 ext{ g mol}^{-1}} \\ = 0.0129 ext{ mol}

Step 2

Determine the moles of sodium phosphate

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Answer

The reaction can be understood as:

ext{Na}_3 ext{PO}_4 + (NH₄)_2MoO₄ \rightarrow (NH₄)_3PO₄·12MoO₃

This indicates that 1 mole of sodium phosphate produces 1 mole of the precipitate. Therefore, the moles of sodium phosphate are also:

0.0129 ext{ mol}

Step 3

Calculate the mass of sodium phosphate

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Answer

To get the mass of sodium phosphate, we will use the molar mass of sodium phosphate (Na₃PO₄), which is approximately 163.94 g mol⁻¹:

ext{Mass} = ext{Moles} imes ext{Molar Mass} \\ = 0.0129 ext{ mol} imes 163.94 ext{ g mol}^{-1} \\ = 2.115 ext{ g}

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