To determine the number of grams of diphosphorus trioxide ([tex]\(P_2O_3\)[/tex]) required to react completely with 4.0 moles of water ([tex]\(H_2O\)[/tex]), we need to follow these steps:
### Step 1: Identify the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
[tex]\[
P_2O_3 + 3H_2O \rightarrow 2H_3PO_3
\][/tex]
### Step 2: Determine the Molar Ratios from the Balanced Equation
From the balanced equation, we see that:
- 1 mole of [tex]\(P_2O_3\)[/tex] reacts with 3 moles of [tex]\(H_2O\)[/tex].
### Step 3: Convert Moles of [tex]\(H_2O\)[/tex] to Moles of [tex]\(P_2O_3\)[/tex]
Given that we have 4.0 moles of [tex]\(H_2O\)[/tex] and using the molar ratio from the balanced equation, we can determine the number of moles of [tex]\(P_2O_3\)[/tex] needed:
[tex]\[
\text{moles of } P_2O_3 = \left( \frac{4.0 \text{ moles } H_2O}{3 \text{ moles } H_2O \text{ per 1 mole } P_2O_3} \right) = \frac{4.0}{3} \approx 1.333 \text{ moles of } P_2O_3
\][/tex]
### Step 4: Convert Moles of [tex]\(P_2O_3\)[/tex] to Grams
We now convert the moles of [tex]\(P_2O_3\)[/tex] to grams using its molar mass. The molar mass of [tex]\(P_2O_3\)[/tex] is given as 109.94 g/mol:
[tex]\[
\text{mass of } P_2O_3= \text{moles of } P_2O_3 \times \text{molar mass of } P_2O_3
\][/tex]
[tex]\[
\text{mass of } P_2O_3 = 1.333 \text{ moles } \times 109.94 \text{ g/mol} \approx 146.59 \text{ g}
\][/tex]
### Conclusion
Thus, the number of grams of diphosphorus trioxide required to react completely with 4.0 moles of water is approximately 146.59 grams.
