Question

checking comprehension
14 multiple - choice question
balance the following equations:
when the equation ( h_3po_4+koh
ightarrow k_3po_4 + h_2o ) is balanced the coefficient in front of the

1. ( p + o_2

ightarrow p_2o_5 )
water is:

1. ( agp+cucl_2

ightarrow agcl + cup_2 )

1. ( na+o_2

ightarrow na_2o )

1. ( h_3po_4+koh

ightarrow k_3po_4+h_2o )

1. ( al_2o_3

ightarrow al+o_2 )

1. ( fe + cl_2

ightarrow fecl_3 )

Explanation:

Step1: Balance the equation

$$\text{H}_3\text{PO}_4+\text{KOH}\to\text{K}_3\text{PO}_4+\text{H}_2\text{O}$$
First, balance the potassium (\(K\)) atoms. There are \(3\) \(K\) atoms on the right - hand side (\(\text{K}_3\text{PO}_4\)), so we put a coefficient of \(3\) in front of \(\text{KOH}\):
$$\text{H}_3\text{PO}_4 + 3\text{KOH}\to\text{K}_3\text{PO}_4+\text{H}_2\text{O}$$

Step2: Balance the hydrogen (\(H\)) and oxygen (\(O\)) atoms

On the left - hand side, the number of \(H\) atoms is \(3 + 3=6\) (from \(\text{H}_3\text{PO}_4\) and \(3\text{KOH}\)), and the number of \(O\) atoms (excluding those in \(\text{PO}_4^{3 -}\)) is \(3\) (from \(3\text{KOH}\)).
On the right - hand side, in \(\text{K}_3\text{PO}_4\) there are no extra \(H\) and \(O\) atoms (excluding \(\text{PO}_4^{3 -}\)). Let the coefficient of \(\text{H}_2\text{O}\) be \(x\).
The number of \(H\) atoms: \(2x\) (from \(\text{H}_2\text{O}\)). The number of \(O\) atoms (excluding \(\text{PO}_4^{3 -}\)): \(x\) (from \(\text{H}_2\text{O}\)).
Since the number of \(H\) atoms: \(2x = 6\), so \(x = 3\).
The balanced equation is \(\text{H}_3\text{PO}_4+3\text{KOH}=\text{K}_3\text{PO}_4 + 3\text{H}_2\text{O}\)

Answer:

3