Question

there are $1.2 \times 10^{24}$ atoms of carbon in a sample of aluminum acetate, $\text{al}(\text{c}_2\text{h}_3\text{o}_2)_3$. how many moles of h are in the sample?

• 3.0 moles h
• 12 moles h
• 1.8 moles h
• 53 moles h

Explanation:

Step1: Find moles of C atoms

Using Avogadro's number ($N_A = 6.022\times10^{23}\ \text{atoms/mol}$), moles of C ($n_C$) is calculated as:
$n_C=\frac{\text{Number of C atoms}}{N_A}=\frac{1.2\times10^{24}\ \text{atoms}}{6.022\times10^{23}\ \text{atoms/mol}}\approx2.0\ \text{mol}$

Step2: Determine C:H mole ratio in $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$

In $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$, each formula unit has $2\times3 = 6$ C atoms and $3\times3 = 9$ H atoms.
Thus, the mole ratio of C to H is $6:9 = 2:3$.

Step3: Calculate moles of H

Let $n_H$ be moles of H. From the ratio $\frac{n_C}{n_H}=\frac{2}{3}$, solve for $n_H$:
$n_H=\frac{3}{2}\times n_C=\frac{3}{2}\times2.0\ \text{mol}=3.0\ \text{mol}$ (Wait, correction: Wait, in $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$, the subscript for $\text{C}_2\text{H}_3\text{O}_2$ is 3, so C atoms: $2\times3 = 6$, H atoms: $3\times3 = 9$? Wait no, original formula: $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$: $\text{C}_2\text{H}_3\text{O}_2^-$ is acetate ion. So per acetate ion: 2 C, 3 H, 2 O. Three acetate ions: 6 C, 9 H, 6 O. Wait, earlier step1: $1.2\times10^{24}$ C atoms. Moles of C: $\frac{1.2\times10^{24}}{6.022\times10^{23}}\approx2.0\ \text{mol}$ (since $6.022\times10^{23}\times2 = 1.2044\times10^{24}$). Then, mole ratio of C to H: 6 C : 9 H = 2 C : 3 H. So if $n_C = 2.0\ \text{mol}$, then $n_H=\frac{9}{6}\times n_C=\frac{3}{2}\times2.0 = 3.0\ \text{mol}$? Wait, but wait—wait, no, wait the formula: $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$: let's recheck the formula. Wait, $\text{C}_2\text{H}_3\text{O}_2^-$: 2 C, 3 H, 2 O. Three of them: 6 C, 9 H, 6 O. So C:H ratio is 6:9 = 2:3. So moles of C is 2.0 mol (from $1.2\times10^{24}$ atoms: $1.2e24 / 6.022e23 ≈ 2.0$). Then moles of H: (9/6)*2.0 = 3.0 mol. Wait, but wait, maybe I miscalculated the ratio. Wait, no—wait, the user's options include 3.0 moles H. Wait, but let's re-express:

Wait, $1.2\times10^{24}$ C atoms. Moles of C: $1.2e24 / 6.022e23 ≈ 2.0$ mol (since $6.022e23 * 2 = 1.2044e24$). Then, in $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$, the number of C atoms per formula unit is 6, and H atoms is 9. So the mole ratio of C to H is 6:9 = 2:3. So moles of H = (9/6) moles of C = (3/2)2.0 = 3.0 mol. So the answer is 3.0 moles H.

Answer:

3.0 moles H (Option: 3.0 moles H)