Salt cake `(Na_(2)SO_(4))` is prepared as follows:
`2NaCl + H_(2) SO_(4) rarr Na_(2)SO_(4) + 2HCl`
How much salt cake could be produced from `100.0 g` of 90% pure saltin the above reaction?

Options

A. `109.8 g`
B. `54.9 g`
C. `36.6 g`
D. `209.8 g`

Explanation

To solve the problem of how much salt cake (Na₂SO₄) can be produced from 100.0 g of 90% pure salt (NaCl), we will follow these steps:

### Step 1: Determine the mass of pure NaCl
Given that we have 100.0 g of 90% pure NaCl, we first need to calculate the mass of pure NaCl in this sample.

\[ \text{Mass of pure NaCl} = 100.0 \, \text{g} \times 0.90 = 90.0 \, \text{g} \]

### Step 2: Calculate the number of moles of NaCl
Next, we need to convert the mass of pure NaCl into moles. The molar mass of NaCl is approximately 58.5 g/mol.

\[ \text{Number of moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{90.0 \, \text{g}}{58.5 \, \text{g/mol}} \approx 1.54 \, \text{moles} \]

### Step 3: Use stoichiometry to find moles of Na₂SO₄ produced
From the balanced chemical equation:

\[ 2 \, \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{HCl} \]

We see that 2 moles of NaCl produce 1 mole of Na₂SO₄. Therefore, we can find the number of moles of Na₂SO₄ produced from the moles of NaCl.

\[ \text{Number of moles of Na₂SO₄} = \frac{1.54 \, \text{moles NaCl}}{2} = 0.77 \, \text{moles Na₂SO₄} \]

### Step 4: Calculate the mass of Na₂SO₄ produced
Now, we will calculate the mass of Na₂SO₄ produced using its molar mass. The molar mass of Na₂SO₄ is approximately 142 g/mol.

\[ \text{Mass of Na₂SO₄} = \text{Number of moles} \times \text{Molar mass} = 0.77 \, \text{moles} \times 142 \, \text{g/mol} \approx 109.5 \, \text{g} \]

### Final Answer
The mass of salt cake (Na₂SO₄) that could be produced from 100.0 g of 90% pure NaCl is approximately **109.5 g**.

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