Complete and balance the following reactions:
$(i){H_2}(g) + Mo{O_3}(s)\xrightarrow{\Delta }$
$(ii)CO(g) + {H_2}(g)\xrightarrow[{catalyst}]{\Delta }$
$(iii){C_3}{H_8}(g) + {H_2}\xrightarrow[{catalyst}]{\Delta }$
$(iv)Zn(s) + NaOH(aq)\xrightarrow{{heat}}$

Hint: The chemical equations can be balanced if we know the complete equation by matching the number of elements on both sides of the arrow. Both sides of the arrow should have an equal number of elements.

Complete step by step answer:
In this question, first we will complete the reaction and then will balance the number of elements on both sides of the reaction.
$(i)$. When hydrogen reacts with a metal oxide then the metal oxide is reduced to corresponding metal and water. Therefore, when molybdenum trioxide ($Mo{O_3}$) reacts with hydrogen (${H_2}$) then molybdenum and water is produced. This reaction can be shown as:
${H_2}(g) + Mo{O_3}(s)\xrightarrow{\Delta }Mo + {H_2}O$
Therefore, the balanced chemical equation is:
$3{H_2}(g) + Mo{O_3}(s)\xrightarrow{\Delta }Mo + 3{H_2}O$
$(ii)$. When carbon monoxide ($CO$) with hydrogen then methanol ($C{H_3}OH$) is formed in the products. This reaction can be represented as :
$CO(g) + {H_2}(g)\xrightarrow[{catalyst}]{\Delta }C{H_3}OH$
As at left side there are two hydrogen and right side there is four hydrogen so we will multiply hydrogen at left side with $2$. The balanced chemical equation can be represented as:
$CO(g) + 2{H_2}(g)\xrightarrow[{catalyst}]{\Delta }C{H_3}OH$
$(iii)$. When propane reacts with hydrogen then the product obtained is ethane. This reaction can be represented as:
${C_3}{H_8} + {H_2} \to {C_2}{H_6}$
Here, at left there are three carbon and on the right side there are two carbon so we will multiply the left side of carbon with $2$ and the right side of carbon with $3$. Therefore, the balanced equation can be written as:
$2{C_3}{H_8} + {H_2} \to 3{C_2}{H_6}$
$(iv)$. When zinc reacts with sodium hydroxide the sodium zincate ($N{a_2}Zn{O_2}$) is formed along with hydrogen gas in the product . This reaction can be written as:
$Zn(s) + NaOH(aq)\xrightarrow{{heat}}N{a_2}Zn{O_2} + {H_2}$
As at left side we have one sodium and at right side we have two sodium so we will multiply sodium hydroxide on left side with $2$. Now, the balanced equation is:
$Zn(s) + 2NaOH(aq)\xrightarrow{{heat}}N{a_2}Zn{O_2} + {H_2}$

Note: Remember that When metal oxides are reacted with hydrogen then hydrogen reduces the metal oxide to metal and water but in some cases like $N{a_2}O,ZnO,{K_2}O$ cannot be reduced by hydrogen.