Question

1. (\text{li}_{3}\text{po}_{4}\to\text{li}+\text{p}+\text{o}_{2}) 7. (\text{v}_{2}\text{o}_{5}+\text{cas}\to\text{cao}+\text{v}_{2}\text{s}_{5}) 8. (\text{nh}_{3}+\text{o}_{2}\to\text{n}_{2}+\text{h}_{2}\text{o})

Explanation:

Step1: Balance equation 6

Balance Li, P and O atoms. For $Li_3PO_4
ightarrow Li + P+O_2$, we need 4 moles of $Li_3PO_4$ to get 12 moles of Li, 4 moles of P and 16 moles of O. On the right - hand side, we can then have 12 moles of Li, 4 moles of P and 8 moles of $O_2$. So the coefficients are 4, 12, 4, 8.

Step2: Balance equation 7

For $V_2O_5+CaS
ightarrow CaO + V_2S_5$, balance the S and O atoms. We need 5 moles of CaS to balance the S atoms and 5 moles of CaO to balance the Ca and O atoms. So the coefficients are 1, 5, 5, 1.

Step3: Balance equation 8

For $NH_3+O_2
ightarrow N_2 + H_2O$, first balance the N atoms. Then balance the H atoms and finally the O atoms. We need 4 moles of $NH_3$, 3 moles of $O_2$ to get 2 moles of $N_2$ and 6 moles of $H_2O$. So the coefficients are 4, 3, 2, 6.

Answer:

1. 4, 12, 4, 8
2. 1, 5, 5, 1
3. 4, 3, 2, 6