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To determine the original mass of the potassium phosphate octahydrate ([tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex]) sample that released 7.93 grams of water upon heating, follow these steps:
1. Determine the molar masses:
- The molar mass of water ([tex]\(H_2O\)[/tex]) is 18.015 grams per mole.
- The molar mass of potassium phosphate ([tex]\(K_3PO_4\)[/tex]) is calculated as follows:
- Potassium (K): [tex]\(3 \times 39.10 \, \text{g/mol} = 117.30 \, \text{g/mol}\)[/tex]
- Phosphorus (P): [tex]\(1 \times 30.97 \, \text{g/mol} = 30.97 \, \text{g/mol}\)[/tex]
- Oxygen (O): [tex]\(4 \times 16.00 \, \text{g/mol} = 64.00 \, \text{g/mol}\)[/tex]
- Total molar mass of [tex]\(K_3PO_4\)[/tex]: [tex]\(117.30 + 30.97 + 64.00 = 212.27 \, \text{g/mol}\)[/tex]
- The molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex] (octahydrate) includes both the molar mass of [tex]\(K_3PO_4\)[/tex] and 8 moles of water:
- Molar mass of [tex]\(K_3PO_4\)[/tex]: 212.27 g/mol
- Molar mass of [tex]\(8 \, H_2O\)[/tex]: [tex]\(8 \times 18.015 \, \text{g/mol} = 144.12 \, \text{g/mol}\)[/tex]
- Total molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex]: [tex]\(212.27 + 144.12 = 356.39 \, \text{g/mol}\)[/tex]
2. Calculate the moles of water released:
- Given mass of water released: 7.93 grams
- Moles of water released [tex]\(\left( \text{mol} \right)\)[/tex]:
[tex]\[ \frac{7.93 \, \text{g}}{18.015 \, \text{g/mol}} = 0.44018873161254507 \text{ moles} \][/tex]
3. Relate the moles of water to the moles of hydrate:
- In [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex], 8 moles of water are associated with 1 mole of hydrate.
- Moles of hydrate [tex]\(\left( \text{mol} \right)\)[/tex]:
[tex]\[ \frac{0.44018873161254507 \text{ moles} \, H_2O}{8} = 0.055023591451568134 \text{ moles} \, K_3PO_4 \cdot 8 H_2O \][/tex]
4. Calculate the original mass of the hydrate:
- Molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex]: 356.39 g/mol
- Mass of the hydrate [tex]\(\left( \text{g} \right)\)[/tex]:
[tex]\[ 0.055023591451568134 \text{ moles} \times 356.39 \text{ g/mol} = 19.609857757424365 \text{ grams} \][/tex]
Thus, the original mass of the potassium phosphate octahydrate sample was approximately 19.61 grams.
1. Determine the molar masses:
- The molar mass of water ([tex]\(H_2O\)[/tex]) is 18.015 grams per mole.
- The molar mass of potassium phosphate ([tex]\(K_3PO_4\)[/tex]) is calculated as follows:
- Potassium (K): [tex]\(3 \times 39.10 \, \text{g/mol} = 117.30 \, \text{g/mol}\)[/tex]
- Phosphorus (P): [tex]\(1 \times 30.97 \, \text{g/mol} = 30.97 \, \text{g/mol}\)[/tex]
- Oxygen (O): [tex]\(4 \times 16.00 \, \text{g/mol} = 64.00 \, \text{g/mol}\)[/tex]
- Total molar mass of [tex]\(K_3PO_4\)[/tex]: [tex]\(117.30 + 30.97 + 64.00 = 212.27 \, \text{g/mol}\)[/tex]
- The molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex] (octahydrate) includes both the molar mass of [tex]\(K_3PO_4\)[/tex] and 8 moles of water:
- Molar mass of [tex]\(K_3PO_4\)[/tex]: 212.27 g/mol
- Molar mass of [tex]\(8 \, H_2O\)[/tex]: [tex]\(8 \times 18.015 \, \text{g/mol} = 144.12 \, \text{g/mol}\)[/tex]
- Total molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex]: [tex]\(212.27 + 144.12 = 356.39 \, \text{g/mol}\)[/tex]
2. Calculate the moles of water released:
- Given mass of water released: 7.93 grams
- Moles of water released [tex]\(\left( \text{mol} \right)\)[/tex]:
[tex]\[ \frac{7.93 \, \text{g}}{18.015 \, \text{g/mol}} = 0.44018873161254507 \text{ moles} \][/tex]
3. Relate the moles of water to the moles of hydrate:
- In [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex], 8 moles of water are associated with 1 mole of hydrate.
- Moles of hydrate [tex]\(\left( \text{mol} \right)\)[/tex]:
[tex]\[ \frac{0.44018873161254507 \text{ moles} \, H_2O}{8} = 0.055023591451568134 \text{ moles} \, K_3PO_4 \cdot 8 H_2O \][/tex]
4. Calculate the original mass of the hydrate:
- Molar mass of [tex]\(K_3PO_4 \cdot 8 H_2O\)[/tex]: 356.39 g/mol
- Mass of the hydrate [tex]\(\left( \text{g} \right)\)[/tex]:
[tex]\[ 0.055023591451568134 \text{ moles} \times 356.39 \text{ g/mol} = 19.609857757424365 \text{ grams} \][/tex]
Thus, the original mass of the potassium phosphate octahydrate sample was approximately 19.61 grams.
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