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

H3PO4

Ca (OH) 2 (s) + H3PO4 (aq)

Ca3 (PO4) 2 (aq) + H2O (1)

balance the equation 1st.

O 6.80 g

O 15.5 g

O 8.60 g

o 13.7 g

O 10.3 g

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