In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 10.0 grams of barium?

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.

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> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.