Match the weight of reactants given in column I with weight of products maked (?) given in column II

	Column I		Column II
a	$\begin{array}{l}2H_2+O_2\stackrel{ }{\to }2H_{2}O\\ 1.0g1.0g?\end{array}$	p	0.56 g
b	$\begin{array}{l}{N}_2+3H_2\stackrel{ }{\to }2N{H}_3\\ 1.0g1.0g?\end{array}$	q	1.333 g
c	$\begin{array}{l}CaC{O}_3\stackrel{\Delta }{\to }CaO+C{O}_2\\ 1.0g?\end{array}$	r	1.125 g
d	$\begin{array}{l}C+2H_2\stackrel{ }{\to }C{H}_4\\ 1.0g1.0g?\end{array}$	s	1.214 g

  $\begin{array}{l}\left(a\to r\right)2H_2+O_2\stackrel{ }{\to }2H_{2}O\\ 1.0g1.0g\\ \frac{1}{2}mol\frac{1}{32}mol\end{array}$
  $O_2$ is the limiting reagent :  $1mol{O}_2\equiv 2mol{H}_{2}O$
 $\Rightarrow \frac{1}{32}mol{O}_2\equiv \frac{1}{16}mol{H}_{2}O=\frac{18}{16}gof{H}_{2}O=1.125g$
 $\begin{array}{l}\left(b\to s\right)N_2+3H_2\stackrel{ }{\to }2N{H}_3\\ 1.0g1.0g\\ \frac{1}{28}mol\frac{1}{2}mol\end{array}$
 $N_2$ is the limiting reagent,
 $\therefore \frac{1}{28}mol{N}_2\equiv \frac{1}{14}molN{H}_3=\frac{1}{14}\times 17g=1.214g$
 $\begin{array}{l}\left(c\to p\right)CaC{O}_3\stackrel{\Delta }{\to }CaO+C{O}_2\\ 1.0g\\ \frac{1}{100}mol\frac{1}{100}mol\frac{1}{100}mol\\ =\frac{1}{100}\times 56gCaO\\ =0.56g\end{array}$
 $\begin{array}{l}\left(d\to q\right)C+2H_2\stackrel{ }{\to }C{H}_4\\ 1.0g1.0g\\ \frac{1}{12}mol\frac{1}{2}mol\end{array}$
C is the limiting reagent.
$\therefore \frac{1}{12}molC\equiv \frac{1}{12}molC{H}_4\equiv \frac{16}{12}g=1.333g$