How many of `P_4`can be producedby reaction of `0.10` moles `Ca_5(PO_4)_3F,0.36` moles `SiO_2 and 0.90` moles `C` according to the following reaction ? `4Ca_5(PO_4)_3F+18 SiO_2+30Crarr3P_4+2CaF_2+18CaSiO_3+30CO`

Options

A. `0.060`
B. `0.030`
C. `0.045`
D. `0.075`

Answer

To determine how many moles of \( P_4 \) can be produced from the reaction of \( 0.10 \) moles of \( Ca_5(PO_4)_3F \), \( 0.36 \) moles of \( SiO_2 \), and \( 0.90 \) moles of \( C \), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ 4 Ca_5(PO_4)_3F + 18 SiO_2 + 30 C \rightarrow 3 P_4 + 2 CaF_2 + 18 CaSiO_3 + 30 CO \]

### Step 2: Identify the Stoichiometric Ratios
From the balanced equation, we can see the stoichiometric ratios:
- \( 4 \) moles of \( Ca_5(PO_4)_3F \) produce \( 3 \) moles of \( P_4 \)
- \( 18 \) moles of \( SiO_2 \) produce \( 3 \) moles of \( P_4 \)
- \( 30 \) moles of \( C \) produce \( 3 \) moles of \( P_4 \)

### Step 3: Calculate the Required Moles of Each Reactant
To find out which reactant is the limiting reagent, we need to calculate how much of each reactant is required to produce \( P_4 \).

1. **For \( Ca_5(PO_4)_3F \)**:
- \( 4 \) moles of \( Ca_5(PO_4)_3F \) require \( 3 \) moles of \( P_4 \)
- Therefore, \( 1 \) mole of \( Ca_5(PO_4)_3F \) produces \( \frac{3}{4} \) moles of \( P_4 \)
- \( 0.10 \) moles of \( Ca_5(PO_4)_3F \) will produce:
\[ 0.10 \times \frac{3}{4} = 0.075 \text{ moles of } P_4 \]

2. **For \( SiO_2 \)**:
- \( 18 \) moles of \( SiO_2 \) require \( 3 \) moles of \( P_4 \)
- Therefore, \( 1 \) mole of \( SiO_2 \) produces \( \frac{3}{18} = \frac{1}{6} \) moles of \( P_4 \)
- \( 0.36 \) moles of \( SiO_2 \) will produce:
\[ 0.36 \times \frac{1}{6} = 0.06 \text{ moles of } P_4 \]

3. **For \( C \)**:
- \( 30 \) moles of \( C \) require \( 3 \) moles of \( P_4 \)
- Therefore, \( 1 \) mole of \( C \) produces \( \frac{3}{30} = \frac{1}{10} \) moles of \( P_4 \)
- \( 0.90 \) moles of \( C \) will produce:
\[ 0.90 \times \frac{1}{10} = 0.09 \text{ moles of } P_4 \]

### Step 4: Determine the Limiting Reagent
Now we compare the amounts of \( P_4 \) produced by each reactant:
- From \( Ca_5(PO_4)_3F \): \( 0.075 \) moles
- From \( SiO_2 \): \( 0.06 \) moles
- From \( C \): \( 0.09 \) moles

The limiting reagent is the one that produces the least amount of \( P_4 \), which is \( SiO_2 \) with \( 0.06 \) moles.

### Step 5: Conclusion
Thus, the maximum amount of \( P_4 \) that can be produced from the reaction is:
\[ \text{Moles of } P_4 = 0.06 \]