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Select the correct answer. Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?Use the periodic table and the polyatomic ion resource. A. 353.59 B. 379.49 402.29 D. 863.49 E. 1,060 g

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Select the correct answer.
Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium
phosphate, how much calcium phosphate can you make?Use the periodic table and the polyatomic ion resource.
A. 353.59
B. 379.49
402.29
D. 863.49
E. 1,060 g

Select the correct answer. Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?Use the periodic table and the polyatomic ion resource. A. 353.59 B. 379.49 402.29 D. 863.49 E. 1,060 g

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JoshuaMaster · Tutor for 5 years

Answer

A. 353.59

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To solve this stoichiometry problem, we need to use the balanced chemical equation for the reaction between sodium phosphate (Na₃PO₄) and calcium chloride (CaCl₂). The balanced equation is:<br /><br />\[ 2 \text{Na}_3\text{PO}_4 + 3 \text{CaCl}_2 \rightarrow 6 \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2 \]<br /><br />Next, we convert the mass of CaCl₂ to moles, use the mole ratio from the balanced equation to find moles of Ca₃(PO₄)₂, and then convert moles of Ca₃(PO₄)₂ to grams.<br /><br />1. Calculate moles of CaCl₂:<br /> \[ \text{Molar mass of CaCl}_2 = 40.08 + 2 \times 35.45 = 110.98 \, \text{g/mol} \]<br /> \[ \text{Moles of CaCl}_2 = \frac{379.4 \, \text{g}}{110.98 \, \text{g/mol}} \approx 3.42 \, \text{mol} \]<br /><br />2. Use the mole ratio to find moles of Ca₃(PO₄)₂:<br /> \[ \text{Mole ratio of CaCl}_2 \text{ to Ca}_3(\text{PO}_4)_2 = \frac{1 \, \text{mol Ca}_3(\text{PO}_4)_2}{3 \, \text{mol CaCl}_2} \]<br /> \[ \text{Moles of Ca}_3(\text{PO}_4)_2 = 3.42 \, \text{mol CaCl}_2 \times \frac{1 \, \text{mol Ca}_3(\text{PO}_4)_2}{3 \, \text{mol CaCl}_2} \approx 1.14 \, \text{mol Ca}_3(\text{PO}_4)_2 \]<br /><br />3. Convert moles of Ca₃(PO₄)₂ to grams:<br /> \[ \text{Molar mass of Ca}_3(\text{PO}_4)_2 = 3 \times 40.08 + 2 \times (30.97 + 4 \times 16.00) = 310.18 \, \text{g/mol} \]<br /> \[ \text{Mass of Ca}_3(\text{PO}_4)_2 = 1.14 \, \text{mol} \times 310.18 \, \text{g/mol} \approx 353.59 \, \text{g} \]
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