Calculate the amount of calcium oxide required when it reacts with `852 g` of `P_(4) O_(10)`.

Answer

To calculate the amount of calcium oxide (CaO) required to react with 852 grams of phosphorus pentoxide (P₄O₁₀), we can follow these steps:

### Step 1: Write the balanced chemical equation

The reaction between calcium oxide and phosphorus pentoxide can be represented as:

\[ \text{CaO} + \text{P}_4\text{O}_{10} \rightarrow \text{Ca}_3\text{(PO}_4\text{)}_2 \]

To balance this equation, we find that:

\[ 6 \text{CaO} + \text{P}_4\text{O}_{10} \rightarrow 2 \text{Ca}_3\text{(PO}_4\text{)}_2 \]

This indicates that 1 mole of P₄O₁₀ reacts with 6 moles of CaO.

### Step 2: Calculate the number of moles of P₄O₁₀

To find the number of moles of P₄O₁₀, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

The molar mass of P₄O₁₀ is calculated as follows:

- Phosphorus (P) = 30.97 g/mol (4 atoms)
- Oxygen (O) = 16.00 g/mol (10 atoms)

Calculating the molar mass:

\[ \text{Molar mass of P}_4\text{O}_{10} = (4 \times 30.97) + (10 \times 16.00) = 123.88 + 160.00 = 283.88 \text{ g/mol} \]

Now, we can calculate the number of moles of P₄O₁₀:

\[ \text{Number of moles of P}_4\text{O}_{10} = \frac{852 \text{ g}}{283.88 \text{ g/mol}} \approx 3.00 \text{ moles} \]

### Step 3: Calculate the number of moles of CaO required

From the balanced equation, we know that 1 mole of P₄O₁₀ requires 6 moles of CaO. Therefore, the number of moles of CaO required is:

\[ \text{Number of moles of CaO} = 6 \times \text{Number of moles of P}_4\text{O}_{10} = 6 \times 3.00 = 18 \text{ moles} \]

### Step 4: Calculate the mass of CaO required

Now we need to find the mass of CaO required using the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

The molar mass of CaO is:

- Calcium (Ca) = 40.08 g/mol
- Oxygen (O) = 16.00 g/mol

Calculating the molar mass of CaO:

\[ \text{Molar mass of CaO} = 40.08 + 16.00 = 56.08 \text{ g/mol} \]

Now, we can calculate the mass of CaO:

\[ \text{Mass of CaO} = 18 \text{ moles} \times 56.08 \text{ g/mol} \approx 1009.44 \text{ g} \]

### Final Answer

The amount of calcium oxide required is approximately **1009.44 grams**.

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