Question

refer to the slides. 2 sig figs molar mass of alcl3 is 133.33 g/mol molar mass of hcl is 36.46 g/mol problem how many moles of hcl will be produced when 249 g of alcl3 are reacted according to this chemical equation? 2alcl3 + 3h2o(ℓ) → al2o3 + 6hcl(g)

Explanation:

Step1: Calculate moles of \( \text{AlCl}_3 \)

Moles of \( \text{AlCl}_3 = \frac{\text{mass of } \text{AlCl}_3}{\text{molar mass of } \text{AlCl}_3} \)
Given mass of \( \text{AlCl}_3 = 249 \, \text{g} \), molar mass of \( \text{AlCl}_3 = 133.33 \, \text{g/mol} \)
\( \text{Moles of } \text{AlCl}_3 = \frac{249}{133.33} \approx 1.8676 \, \text{mol} \)

Step2: Use stoichiometry to find moles of \( \text{HCl} \)

From the balanced equation \( 2\text{AlCl}_3 + 3\text{H}_2\text{O}(l)
ightarrow \text{Al}_2\text{O}_3 + 6\text{HCl}(g) \), the mole ratio of \( \text{AlCl}_3 \) to \( \text{HCl} \) is \( 2:6 \) (or \( 1:3 \)).
Let \( n_{\text{HCl}} \) be moles of \( \text{HCl} \).
\( \frac{n_{\text{HCl}}}{n_{\text{AlCl}_3}} = \frac{6}{2} = 3 \)
\( n_{\text{HCl}} = 3 \times n_{\text{AlCl}_3} \)
Substitute \( n_{\text{AlCl}_3} \approx 1.8676 \, \text{mol} \)
\( n_{\text{HCl}} = 3 \times 1.8676 \approx 5.6028 \, \text{mol} \)

Step3: Round to 2 significant figures

\( 5.6028 \, \text{mol} \approx 5.6 \, \text{mol} \) (or check if 2 sig figs: 249 has 3, 133.33 is more, so result should have 2? Wait, 249 is 3 sig figs, but the problem says "2 sig figs" in the prompt. Wait, maybe the given mass is 249 g (3 sig figs), but the problem says "2 sig figs" in the instruction. Wait, let's re - check. The problem statement says "2 sig figs" at the top. So we need to round to 2 sig figs. 5.6028 rounded to 2 sig figs is 5.6? Wait, no: 5.6028, first two sig figs are 5 and 6, the next digit is 0, so 5.6? Wait, no, 5.6028 is approximately 5.6 when rounded to 2 significant figures? Wait, no, 5.6028: the first significant figure is 5, second is 6, third is 0. So when rounding to 2 sig figs, it's 5.6? Wait, no, 5.6028 is 5.6 when rounded to two significant figures? Wait, actually, 5.6028: the number is 5.6028. The first two significant digits are 5 and 6. The next digit is 0, which is less than 5, so we keep the second digit as is. So 5.6 mol? Wait, but let's check the calculation again.

Wait, moles of \( \text{AlCl}_3=\frac{249}{133.33}\approx1.8676 \). Then moles of \( \text{HCl}=1.8676\times\frac{6}{2}=1.8676\times3 = 5.6028 \). Now, if we consider significant figures: the mass of \( \text{AlCl}_3 \) is 249 g (3 sig figs), molar mass of \( \text{AlCl}_3 \) is 133.33 (5 sig figs), so the moles of \( \text{AlCl}_3 \) should have 3 sig figs. Then the mole ratio is exact (from the balanced equation), so the moles of \( \text{HCl} \) should have 3 sig figs. But the problem says "2 sig figs" in the prompt. So we round 5.6028 to 2 sig figs, which is 5.6? Wait, no, 5.6028 rounded to two significant figures: the first two are 5 and 6, the third digit is 0, so it's 5.6? Wait, no, 5.6028 is 5.6 when rounded to two significant figures? Wait, actually, 5.6028: the first significant figure is 5, the second is 6, the third is 0. So when rounding to two significant figures, we look at the third digit. Since 0 < 5, we don't round up the second digit. So it's 5.6 mol. But wait, maybe I made a mistake in the mole ratio. Let's check the balanced equation again: \( 2\text{AlCl}_3 \) produces \( 6\text{HCl} \), so the ratio of \( \text{AlCl}_3 \) to \( \text{HCl} \) is \( 2:6 = 1:3 \). So for every 1 mole of \( \text{AlCl}_3 \), we get 3 moles of \( \text{HCl} \). So moles of \( \text{AlCl}_3=\frac{249}{133.33}\approx1.8676 \), moles of \( \text{HCl}=1.8676\times3 = 5.6028 \). Rounding to 2 significant figures: 5.6 mol? Wait, 5.6028 is closer to 5.6 than 5.7? Wait, no, 5.6028 is 5.6 when rounded to two significant figures because the third digit is 0.

Answer:

\( \boxed{5.6} \) (in moles)