Question

Match the weight of reactants given in column I with weight of products marked given in column II.

Answer 1

$A:2H_2+O_2\to 2H_{2}O$

Here $O_2$ is the limiting reagent, as $32$ g of $O_2$ gives $36$ g of $H_{2}O$.
So, $1$ g will give $=\frac{1\times 36}{32}=1.125$ g of $H_{2}O$

$B:N_2+3H_2\to 2N{H}_3$

Here $N_2$ is the limiting reagent, as $28$ g of $N_2$ gives $34$ g of $N{H}_3$
So, $1$ g of $N_2$ will give $=\frac{1\times 34}{28}=1.214$ g. of $N{H}_3$.

$C:CaC{O}_3\to CaO+C{O}_2$

As $100$ g of $CaC{O}_3$ gives $56$ g of $CaO$, so $1$ g of $CaC{O}_3$ will give the $0.56$ g of $CaO$.

$D:C+2H_2\to C{H}_4$

$C$ is limiting reagent as $12$ g of $C$ gives $16$ g of $C{H}_4$.
So, $1$ g of $C$ will give $=\frac{1\times 16}{12}=1.333$ g of $C{H}_4$.