part c\n2.20×10⁻² mol h₂co₃\nview available hint(s)\nmoles of oxygen = mol\nsubmit\npart d\n22.7 mol…

Explanation:

Step1: Determine O atoms in $H_2CO_3$

In one molecule of $H_2CO_3$, there are 3 oxygen atoms. So the mole - ratio of $H_2CO_3$ to $O$ is 1:3.

Step2: Calculate moles of O in $H_2CO_3$

Given $n_{H_2CO_3}=2.20\times 10^{-2}\ mol$. Then $n_O = 3\times n_{H_2CO_3}=3\times2.20\times 10^{-2}\ mol = 6.60\times 10^{-2}\ mol$.

Step3: Determine O atoms in $CO_2$

In one molecule of $CO_2$, there are 2 oxygen atoms. So the mole - ratio of $CO_2$ to $O$ is 1:2.

Step4: Calculate moles of O in $CO_2$

Given $n_{CO_2}=22.7\ mol$. Then $n_O = 2\times n_{CO_2}=2\times22.7\ mol = 45.4\ mol$.

Answer:

Part C: $6.60\times 10^{-2}$ Part D: $45.4$