Question

1. ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + h₂o

4 fes + o → feo + so

Explanation:

Step1: Balance Ca atoms

On the left - hand side, there is 1 Ca atom in \(Ca(OH)_2\), and on the right - hand side, there are 3 Ca atoms in \(Ca_3(PO_4)_2\). So, we put a coefficient of 3 in front of \(Ca(OH)_2\).

$$3Ca(OH)_2 + H_3PO_4 ightarrow Ca_3(PO_4)_2+H_2O$$

Step2: Balance \(PO_4\) groups

On the left - hand side, there is 1 \(H_3PO_4\) (1 \(PO_4\) group), and on the right - hand side, there are 2 \(PO_4\) groups in \(Ca_3(PO_4)_2\). So, we put a coefficient of 2 in front of \(H_3PO_4\).

$$3Ca(OH)_2 + 2H_3PO_4 ightarrow Ca_3(PO_4)_2+H_2O$$

Step3: Balance H and O atoms

On the left - hand side, the number of H atoms: \(3\times2 + 2\times3=12\), and the number of O atoms: \(3\times2+2\times4 = 14\). On the right - hand side, in \(Ca_3(PO_4)_2\), there are \(2\times4 = 8\) O atoms. Let the coefficient of \(H_2O\) be \(x\). For H atoms: \(2x=12\), so \(x = 6\). For O atoms: \(8 + x=14\), \(x = 6\) also satisfies the O - atom balance.

Answer:

$$3Ca(OH)_2 + 2H_3PO_4=Ca_3(PO_4)_2 + 6H_2O$$