$v_{H_{2}}: v_{O_{2}}: v_{C H_{2}}=n_{H_{12}}: n_{O_{2}}: n_{C H_{4}}$
$\Rightarrow v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m_{H_{2}}}{M_{H_{2}}}: \frac{m_{O_{2}}}{M_{O_{2}}}: \frac{m_{C H_{4}}}{M_{C H_{4}}}$
But $m_{H_{2}}=m_{O_{2}}=m_{C H_{4}}=m\left[\therefore n=\frac{\text { mass }}{\text { molar massd }}\right]$
Thus, $\quad v_{H_{2}}: v_{O_{2}}: v_{C H_{4}}=\frac{m}{2}=\frac{m}{1}=\frac{m}{16}=16: 1: 2$
$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે
$2VO + 3Fe_2O_3 → 6FeO + V_2O_5 (V = 51, Fe = 56)$