A sample was contaminated with sodium phosphate. The sample was dissolved in water and added to an excess of acidified (NH₂)₄MoO₄ 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 (NH₂)₄MoO₄ 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 number of moles of the precipitate obtained, we use the formula:

ext{Number of moles} = rac{ ext{mass (g)}}{ ext{molar mass (g/mol)}}

Substituting the values:

ext{Number of moles} = rac{24.21 ext{ g}}{1877 ext{ g/mol}} \ \ = 0.0129 ext{ moles}

Step 2

Determine the moles of sodium phosphate

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Answer

According to the reaction, one mole of sodium phosphate reacts to produce one mole of (NH₄)₃PO₄·12MoO₃. Therefore, the moles of sodium phosphate is also:

= 0.0129 ext{ moles}

Step 3

Calculate the mass of sodium phosphate

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Answer

Using the molar mass of sodium phosphate: (Na₃PO₄) is approximately 163.94 g/mol, the mass of sodium phosphate can be calculated as follows:

ext{Mass} = ext{moles} imes ext{molar mass}

Substituting the values:

ext{Mass} = 0.0129 ext{ moles} imes 163.94 ext{ g/mol} \ \ = 2.115 ext{ g}

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