Question

1. __ h_{3}po_{4}+__ koh

ightarrow __ k_{3}po_{4}+__ h_{2}o

1. k+ b_{2}o_{3}

ightarrow __ k_{2}o+__

1. hcl+ naoh

ightarrow nacl+ h_{2}o

1. na+ nano_{3}

ightarrow __ na_{2}o+__ n_{2}

Explanation:

Step1: Balance the first equation

For \(H_3PO_4+KOH
ightarrow K_3PO_4 + H_2O\), balance \(K\) and \(H\) atoms.
$$H_3PO_4 + 3KOH=K_3PO_4+3H_2O$$

Step2: Balance the second equation

For \(K + B_2O_3
ightarrow K_2O + B\), balance \(K\) and \(B\) atoms.
$$6K + B_2O_3 = 3K_2O+2B$$

Step3: Balance the third equation

For \(HCl+NaOH
ightarrow NaCl + H_2O\), it is a simple acid - base reaction.
$$HCl+NaOH = NaCl+H_2O$$

Step4: Balance the fourth equation

For \(Na+NaNO_3
ightarrow Na_2O+N_2\), use the oxidation - reduction method or trial - and - error.
$$10Na + 2NaNO_3=6Na_2O+N_2$$

Answer:

1. \(1\) \(3\) \(1\) \(3\)
2. \(6\) \(1\) \(3\) \(2B\)
3. \(1\) \(1\) \(1\) \(1\)
4. \(10\) \(2\) \(6\) \(1\)