Question

How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

Answer 1

writhe down the equation first

Answer 2

`H2` reacts with `O2` \[\large\rm H_2 + O_2 \rightarrow H_2O\]
balance the chemical equation
Left side: 2 H and 2 Oxygen
Right side: 2 H, only one oxygen
\[\large\rm H_2 + O_2 \rightarrow \color{red}{2} H_2O\]
Left side: `2` H and `2` Oxygen
Right side: `4 H`, only `2` oxygen
we need 2 more hydrogen on left side
\[\large\rm 2H_2 + O_2 \rightarrow \color{red}{2} H_2O\]
\[\rm 10.54 g ~H_2 (\frac{1 ~mol~H_2}{2.02 ~g~ H_2}) = 5.22 ~mol ~H_2\]\[\rm 95.10 g ~O_2 (\frac{1 ~mol~O_2}{32 ~g~ O_2}) = \color{orange}{2.97}~mol ~O_2\]
Oxygen being the lowest =>it will be the limiting reagent

\[\large\rm 2.97 ~mol~O_2 (\frac{2 ~mol~H_2 O}{1 ~mol~O_2})(\frac{18.02 ~g~H_2O}{1 ~mol~H_2O}) \]\[\ \rm 2.97 ~\cancel{mol~O_2 }(\frac{2 ~\color{Red}{\cancel{mol~H_2 O}}}{\cancel{1 ~mol~O_2}})(\frac{18.02 ~g~H_2O}{\color{red}{\cancel{1 ~mol~H_2O}}}) = ??? g ~H_2O\]