The mass of the precipitate i.e Ba₃(PO₄)₂ obtained from the reaction is 4.82 g
We'll begin by writing the balanced equation for the reaction. This is given below:
3BaCl₂(aq) + 2Na₃PO₄(aq) —> Ba₃(PO₄)₂(s) + 6NaCl(aq)
Next, we shall determine the mass of BaCl₂ that reacted and the mass of Ba₃(PO₄)₂ produced from the balanced equation. This is illustrated below:
Molar mass of BaCl₂ = 137 + (35.5×2)
= 137 + 71
= 208 g/mol
Mass of BaCl₂ from the balanced equation = 3 × 208 = 624 g
Molar mass of Ba₃(PO₄)₂ = (3×137) + 2[31 + (4×16)]
= 411 + 2[31 + 64]
= 411 + 2[95]
= 411 + 109
= 601 g/mol
Mass of Ba₃(PO₄)₂ from the balanced equation = 1 × 601 = 601 g
Thus,
From the balanced equation above,
624 g of BaCl₂ reacted to produce 601 g of Ba₃(PO₄)₂.
Finally, we shall determine the mass of the precipitate i.e Ba₃(PO₄)₂ produced by the reaction of 5 g of BaCl₂. This is illustrated below:
From the balanced equation above,
624 g of BaCl₂ reacted to produce 601 g of Ba₃(PO₄)₂.
Therefore,
5 g of BaCl₂ will react to produce = [tex]\frac{5 * 601}{624}[/tex] = 4.82 g of Ba₃(PO₄)₂.
Thus, 4.82 g of the precipitate i.e Ba₃(PO₄)₂ were obtained from the reaction.
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