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To find out how many grams of [tex]\( H_3PO_3 \)[/tex] would be produced from the complete reaction of 92.3 grams of [tex]\( P_2O_3 \)[/tex], we will follow these steps:
1. Determine the moles of [tex]\( P_2O_3 \)[/tex] in 92.3 grams:
- Given:
- Mass of [tex]\( P_2O_3 \)[/tex] = 92.3 grams
- Molar mass of [tex]\( P_2O_3 \)[/tex] = 109.94 grams per mole
- Formula to find moles:
[tex]\[ \text{Moles of } P_2O_3 = \frac{\text{Mass of } P_2O_3}{\text{Molar mass of } P_2O_3} \][/tex]
- Calculation:
[tex]\[ \text{Moles of } P_2O_3 = \frac{92.3 \, \text{grams}}{109.94 \, \text{grams per mole}} \approx 0.8395 \, \text{moles} \][/tex]
2. Use the balanced chemical equation to find the moles of [tex]\( H_3PO_3 \)[/tex] formed:
- From the balanced chemical equation:
[tex]\[ P_2O_3 + 3 H_2O \rightarrow 2 H_3PO_3 \][/tex]
We see that 1 mole of [tex]\( P_2O_3 \)[/tex] produces 2 moles of [tex]\( H_3PO_3 \)[/tex].
- Using the moles of [tex]\( P_2O_3 \)[/tex] we calculated:
[tex]\[ \text{Moles of } H_3PO_3 = 0.8395 \, \text{moles of } P_2O_3 \times 2 = 1.6791 \, \text{moles} \][/tex]
3. Calculate the mass of [tex]\( H_3PO_3 \)[/tex] produced:
- Given:
- Molar mass of [tex]\( H_3PO_3 \)[/tex] = 82.00 grams per mole
- Formula to find mass:
[tex]\[ \text{Mass of } H_3PO_3 = \text{Moles of } H_3PO_3 \times \text{Molar mass of } H_3PO_3 \][/tex]
- Calculation:
[tex]\[ \text{Mass of } H_3PO_3 = 1.6791 \, \text{moles} \times 82.00 \, \text{grams per mole} \approx 137.69 \, \text{grams} \][/tex]
So, the mass of [tex]\( H_3PO_3 \)[/tex] produced from the complete reaction of 92.3 grams of [tex]\( P_2O_3 \)[/tex] is approximately 137.69 grams.
1. Determine the moles of [tex]\( P_2O_3 \)[/tex] in 92.3 grams:
- Given:
- Mass of [tex]\( P_2O_3 \)[/tex] = 92.3 grams
- Molar mass of [tex]\( P_2O_3 \)[/tex] = 109.94 grams per mole
- Formula to find moles:
[tex]\[ \text{Moles of } P_2O_3 = \frac{\text{Mass of } P_2O_3}{\text{Molar mass of } P_2O_3} \][/tex]
- Calculation:
[tex]\[ \text{Moles of } P_2O_3 = \frac{92.3 \, \text{grams}}{109.94 \, \text{grams per mole}} \approx 0.8395 \, \text{moles} \][/tex]
2. Use the balanced chemical equation to find the moles of [tex]\( H_3PO_3 \)[/tex] formed:
- From the balanced chemical equation:
[tex]\[ P_2O_3 + 3 H_2O \rightarrow 2 H_3PO_3 \][/tex]
We see that 1 mole of [tex]\( P_2O_3 \)[/tex] produces 2 moles of [tex]\( H_3PO_3 \)[/tex].
- Using the moles of [tex]\( P_2O_3 \)[/tex] we calculated:
[tex]\[ \text{Moles of } H_3PO_3 = 0.8395 \, \text{moles of } P_2O_3 \times 2 = 1.6791 \, \text{moles} \][/tex]
3. Calculate the mass of [tex]\( H_3PO_3 \)[/tex] produced:
- Given:
- Molar mass of [tex]\( H_3PO_3 \)[/tex] = 82.00 grams per mole
- Formula to find mass:
[tex]\[ \text{Mass of } H_3PO_3 = \text{Moles of } H_3PO_3 \times \text{Molar mass of } H_3PO_3 \][/tex]
- Calculation:
[tex]\[ \text{Mass of } H_3PO_3 = 1.6791 \, \text{moles} \times 82.00 \, \text{grams per mole} \approx 137.69 \, \text{grams} \][/tex]
So, the mass of [tex]\( H_3PO_3 \)[/tex] produced from the complete reaction of 92.3 grams of [tex]\( P_2O_3 \)[/tex] is approximately 137.69 grams.
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