Question

What is the maximum amount of Ca3(PO4)2 that can be prepared from 9.80 g of Ca(OH)2 and 9.80 g of

Answer 1

Answer:

13.7g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

3Ca(OH)2(s) + 2H3PO4(aq) —> Ca3(PO4)2(aq) + 12H2O(l)

Step 2:

Determination of the masses of Ca(OH)2 and H3PO4 that reacted and the mass of Ca3(PO4)2 produced from the balanced equation.

Molar mass of Ca(OH)2 = 40 + 2(16 + 1) = 74g/mol

Mass of Ca(OH)2 from the balanced equation = 3 x 74 = 222g

Molar mass of H3PO4 = (3x1) + 31 + (16x4) = 98g/mol

Mass of H3PO4 from the balanced equation = 2 x 98 = 196g

Molar mass of Ca3(PO4)2 = (40x3) + 2[31 + (16x4)]

= 120 + 2[95] = 310g/mol

Mass of Ca3(PO4)2 from the balanced equation = 1 x 310 = 310g.

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4 to produce 310g of Ca3(PO4)2.

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4.

Therefore, 9.8g of Ca(OH)2 will react with = (9.8 x 196)/222 = 8.65g of H3PO4

From the above illustration, we can see that only 8.65g of H3PO4 out 9.8g given reacted completely with 9.8g of Ca(OH)2. Therefore, Ca(OH)2 is the limiting reactant and H3PO4 is the excess reactant.

Step 4:

Determination of the maximum mass of Ca3(PO4)2 produced from reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction.

The limiting reactant is Ca(OH)2 and maximum mass of Ca3(PO4)2 produced can be obtained as follow:

From the balanced equation above,

222g of Ca(OH)2 reacted to produce 310g of Ca3(PO4)2.

Therefore, 9.8g of Ca(OH)2 will react to produce = (9.8 x 310)/222 = 13.7g of Ca3(PO4)2.

Therefore, the maximum mass of Ca3(PO4)2 produced is 13.7g