In a determination of P an aqueous solution of `NaH_(2)PO_(4)` is treated with a mixture of ammonium and magnesium ammonium phosphate `Mg(NH_(4))PO_(4).6H_(2)O`. This is heated and decomposed to magnesium pyrophosphate, `Mg_(2)P_(2)O_(7)` which is weighed . A solution of `NaH_(2)PO_(4)` yeilded 1.11 g of `Mg_(2)P_(2)O_(7)`. What weight of `NaH_(2)PO_(4)` was present originally ? (P = 31)

Answer

To solve the problem, we need to determine the weight of `NaH₂PO₄` that was originally present based on the weight of `Mg₂P₂O₇` produced. Here’s a step-by-step solution:

### Step 1: Calculate the molar mass of `Mg₂P₂O₇`
The molar mass of `Mg₂P₂O₇` can be calculated as follows:
- Molar mass of Mg = 24.31 g/mol
- Molar mass of P = 31 g/mol
- Molar mass of O = 16 g/mol

Calculating the total:
- Molar mass of `Mg₂P₂O₇` = (2 × 24.31) + (2 × 31) + (7 × 16)
- Molar mass of `Mg₂P₂O₇` = 48.62 + 62 + 112 = 222.62 g/mol (approximately 222 g/mol)

### Step 2: Determine the amount of phosphorus in `Mg₂P₂O₇`
From the problem, we know that 1.11 g of `Mg₂P₂O₇` was produced. We need to find out how much phosphorus is contained in this mass.

Using the ratio of phosphorus in `Mg₂P₂O₇`:
- In 222 g of `Mg₂P₂O₇`, there are 62 g of phosphorus.
- Therefore, in 1.11 g of `Mg₂P₂O₇`, the amount of phosphorus can be calculated as:

\[ \text{Phosphorus} = \left( \frac{62 \, \text{g}}{222 \, \text{g}} \right) \times 1.11 \, \text{g} = 0.31 \, \text{g} \]

### Step 3: Relate phosphorus to `NaH₂PO₄`
Now, we know that the phosphorus in `Mg₂P₂O₇` comes from `NaH₂PO₄`. The molar mass of `NaH₂PO₄` is:
- Molar mass of Na = 23 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of P = 31 g/mol
- Molar mass of O = 16 g/mol

Calculating the total:
- Molar mass of `NaH₂PO₄` = 23 + (2 × 1) + 31 + (4 × 16) = 120 g/mol

### Step 4: Calculate the amount of `NaH₂PO₄` corresponding to the phosphorus
From the molar mass, we know that 31 g of phosphorus is present in 120 g of `NaH₂PO₄`. We can set up a proportion to find the mass of `NaH₂PO₄` that contains 0.31 g of phosphorus:

\[ \text{Weight of } NaH₂PO₄ = \left( \frac{120 \, \text{g}}{31 \, \text{g}} \right) \times 0.31 \, \text{g} \]

Calculating this gives:

\[ \text{Weight of } NaH₂PO₄ = \frac{120 \times 0.31}{31} = 1.2 \, \text{g} \]

### Final Answer
The weight of `NaH₂PO₄` that was originally present is **1.2 g**.

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