Verify by calculating that:
$(a)$ $5$ moles of $CO_2$ has a higher mass than $5$ moles of $H_2O$.
$(b)$ $120 \, g$ of calcium and $120 \, g$ of magnesium elements have a mole ratio of $3:5$.
$(a)$ $5$ moles of $CO_2$ has a higher mass than $5$ moles of $H_2O$.
$(b)$ $120 \, g$ of calcium and $120 \, g$ of magnesium elements have a mole ratio of $3:5$.
(N/A) The molar mass of $CO_2$ is $44 \, g/mol$ and the molar mass of $H_2O$ is $18 \, g/mol$.
Mass of $5$ moles of $CO_2 = 5 \times 44 = 220 \, g$.
Mass of $5$ moles of $H_2O = 5 \times 18 = 90 \, g$.
Since $220 \, g > 90 \, g$, it is verified that $5$ moles of $CO_2$ has a higher mass.
$(b)$ Molar mass of $Ca = 40 \, g/mol$ and $Mg = 24 \, g/mol$.
Moles of $120 \, g$ of $Ca = 120 / 40 = 3 \, moles$.
Moles of $120 \, g$ of $Mg = 120 / 24 = 5 \, moles$.
Therefore, the mole ratio of $Ca$ to $Mg = 3:5$.
Mass of $5$ moles of $CO_2 = 5 \times 44 = 220 \, g$.
Mass of $5$ moles of $H_2O = 5 \times 18 = 90 \, g$.
Since $220 \, g > 90 \, g$, it is verified that $5$ moles of $CO_2$ has a higher mass.
$(b)$ Molar mass of $Ca = 40 \, g/mol$ and $Mg = 24 \, g/mol$.
Moles of $120 \, g$ of $Ca = 120 / 40 = 3 \, moles$.
Moles of $120 \, g$ of $Mg = 120 / 24 = 5 \, moles$.
Therefore, the mole ratio of $Ca$ to $Mg = 3:5$.
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