Answer
a) From 1.75 moles of $H_{2}$, 3.5 moles of $HCl$ are obtained.
b) From 1.75 moles of $O_{2}$, 3.5 moles of $H_{2}O$ are obtained.
c) From 1.75 moles of $Na$, 0.875 moles of $Na_{2}O_{2}$ are obtained.
d) From 1.75 moles of $H_{2}$, 1.167 moles of $SO_{3}$ are obtained.
Work Step by Step
a) $H_{2} + Cl_{2} = 2HCl$
In this reaction, 2 moles of HCl are produced when 1 mole of $H_{2}$ reacts with $Cl_{2}$.
Therefore for 1.75 moles of $H_{2}$,
$(2\times1.75)$ moles, ie 3.5 moles of HCl are produced.
b) $2H_{2} + O_{2} = 2H_{2}O$
In this reaction, 2 moles of $H_{2}O$ are produced when 1 mole of $O_{2}$ reacts with $H_{2}$.
Therefore for 1.75 moles of $O_{2}$,
$(2\times1.75)$ moles, ie 3.5 moles of $H_{2}O$ are produced.
c) $2Na + O_{2} = Na_{2}O_{2}$
In this reaction, 1 mole of $Na_{2}O_{2}$ is produced when 2 moles of Na reacts with $O_{2}$.
Therefore for 1.75 moles of Na,
$(\frac{1.75}{2})$ moles, ie 0.875 moles of $Na_{2}O_{2}$ are produced.
d) $2S + 3O_{2} = 2SO_{3}$
In this reaction, 2 moles of $SO_{3}$ are produced when 3 moles of $O_{2}$ reacts with S.
Therefore for 1.75 moles of $O_{2}$,
$(\frac{2\times1.75}{3})$ moles, ie 1.167 moles of $SO_{3}$ are produced.
