Question

. When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of Ca3(PO4)2? Ca(OH)2 + H3PO4 yields H2O + Ca3(PO4)2 1 2 3 6

Answer 1

Answer: Option (a) is the correct answer.

Step-by-step explanation:

An equation can only be balanced when number of reactants equal the number of products.

The given equation is as follows.

`Ca(OH)_(2) + H_(3)PO_(4) \rightarrow Ca_(3)(PO_(4))_(2) + H_(2)O`

The number of reactant atoms are as follows.

• Ca = 1
• O = 6
• H = 5
• P = 1

The number of product atoms are as follows.

• Ca = 3
• O = 9
• H = 2
• P = 2

To balance the equation, multiply
`Ca(OH)_(2)` by 3, and
`H_(3)PO_(4)` by 2 on the reactant side. Multiply
`H_(2)O` by 6 on the product side. Therefore, the equation will be as follows.

`3Ca(OH)_(2) + 2H_(3)PO_(4) \rightarrow Ca_(3)(PO_(4))_(2) + 6H_(2)O`

Thus, we can conclude that the coefficient in front of
`Ca_(3)(PO_(4))_(2)` is 1.

Answer 2

2H3PO4 + 3Ca(OH)2 ===>>> 6 H20 + Ca3(PO4)2

Answer
1 <<<<<=====