Answer:
A. 3.36g of Na.
B. 14.62g of Ba3(PO4)2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2
Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.
This is illustrated below:
Molar Mass of Ba = 137g/mol
Mass of Ba from the balanced equation = 3 x 137 = 411g
Molar mass of Na = 23g/mol
Mass of Na from the balanced equation = 6 x 23 = 138g
Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol
Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g
Summary:
From the balanced equation above,
411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.
A. Determination of the mass of Na produced by reacting 10g of Ba.
From the balanced equation above,
411g of Ba reacted to produce 138g of Na.
Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.
Therefore, 3.36g of Na is produced.
B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.
From the balanced equation above,
411g of Ba reacted to produce 601g of Ba3(PO4)2.
Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.
Therefore, 14.62g of Ba3(PO4)2 is produced.
