3CaCl2 + 2Na3PO4 = 6NaCl + Ca3 (PO4) 2 How many grams and moles of Ca3 (PO4) 2 are formed

| April 9, 2021 | education

3CaCl2 + 2Na3PO4 = 6NaCl + Ca3 (PO4) 2 How many grams and moles of Ca3 (PO4) 2 are formed if 3.3 grams of calcium chloride reacts?

Given:
m (CaCl2) = 3.3 g

To find:
n (Ca3 (PO4) 2) -?
m (Ca3 (PO4) 2) -?

Decision:
1) 3CaCl2 + 2Na3PO4 => 6NaCl + Ca3 (PO4) 2;
2) M (CaCl2) = Mr (CaCl2) = Ar (Ca) * N (Ca) + Ar (Cl) * N (Cl) = 40 * 1 + 35.5 * 2 = 111 g / mol;
M (Ca3 (PO4) 2) = Mr (Ca3 (PO4) 2) = Ar (Ca) * N (Ca) + Ar (P) * N (P) + Ar (O) * N (O) = 40 * 3 + 31 * 2 + 16 * 4 * 2 = 310 g / mol;
3) n (CaCl2) = m (CaCl2) / M (CaCl2) = 3.3 / 111 = 0.03 mol;
4) n (Ca3 (PO4) 2) = n (CaCl2) / 3 = 0.03 / 3 = 0.01 mol;
5) m (Ca3 (PO4) 2) = n (Ca3 (PO4) 2) * M (Ca3 (PO4) 2) = 0.01 * 310 = 3.1 g.

Answer: The amount of Ca3 (PO4) 2 substance is 0.01 mol; weight – 3.1 g.



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