Question

1. 2al + 2hcl → 3h₂ + 2alcl₃

al = 2 h = 2
cl = 2
al = 2 h = 6
cl = 6
yes no

1. 2sno₂ + 4h₂ → 2sn + 3h₂o

sn = h =
o =
sn = h =
o =
yes no

1. 2nh₃+ 2o₂ → no + h₂o

n = h =
o =
n = h =
o =
yes no

Explanation:

Step1: Count atoms for 7th reaction - reactant side

On the left - hand side of $2SnO_2 + 4H_2
ightarrow2Sn + 3H_2O$, for $Sn$: coefficient of $Sn$ in $SnO_2$ is 2, so $Sn = 2$; for $H$: coefficient of $H_2$ is 4, so $H=8$; for $O$: coefficient of $O$ in $SnO_2$ is 2 and its sub - script is 2, so $O = 4$.

Step2: Count atoms for 7th reaction - product side

On the right - hand side, for $Sn$: coefficient of $Sn$ is 2, so $Sn = 2$; for $H$: coefficient of $H_2O$ is 3, so $H = 6$; for $O$: coefficient of $O$ in $H_2O$ is 3, so $O=3$. Since the number of $H$ and $O$ atoms are not equal on both sides, the answer is No.

Step3: Count atoms for 9th reaction - reactant side

For $2NH_3+2O_2
ightarrow NO + H_2O$, on the left - hand side, for $N$: coefficient of $N$ in $NH_3$ is 2, so $N = 2$; for $H$: coefficient of $H$ in $NH_3$ is 3 and its coefficient is 2, so $H = 6$; for $O$: coefficient of $O$ in $O_2$ is 2, so $O = 4$.

Step4: Count atoms for 9th reaction - product side

On the right - hand side, for $N$: coefficient of $N$ in $NO$ is 1, so $N = 1$; for $H$: coefficient of $H$ in $H_2O$ is 2, so $H = 2$; for $O$: coefficient of $O$ in $NO$ is 1 and in $H_2O$ is 1, so $O=2$. Since the number of $N$, $H$ and $O$ atoms are not equal on both sides, the answer is No.

Answer:

1. Sn: 2, H: 8, O: 4; Sn: 2, H: 6, O: 3; No
2. N: 2, H: 6, O: 4; N: 1, H: 2, O: 2; No