Question

how many moles of iron are contained in 1.55 kg of iron? give the number with the correct significant figures and unit
question 41 (0.0862 points)
how many moles of aluminum are needed to make 6 moles of h₂?
give just the number with the correct significant figures.
given the reaction: 2 al + 6 hcl → 2 alcl₃ + 3h₂
question 42 (0.0862 points)
what volume in l of 9.00 m nitric acid is needed to make 6.5 l of 1.00 m solution?
give the number with the correct significant figures and unit.

Explanation:

Question (Top, Moles of Iron):

Step1: Convert mass to grams

Given mass of iron is \( 1.55 \, \text{kg} \). Convert to grams: \( 1.55 \, \text{kg} \times 1000 \, \text{g/kg} = 1550 \, \text{g} \).

Step2: Use molar mass of Fe

Molar mass of iron (Fe) is \( 55.85 \, \text{g/mol} \). Moles \( = \frac{\text{mass}}{\text{molar mass}} = \frac{1550 \, \text{g}}{55.85 \, \text{g/mol}} \approx 27.75 \, \text{mol} \). With significant figures (3 from 1.55 kg), it's \( 27.8 \, \text{mol} \) (or more accurately, \( \approx 27.7 \, \text{mol} \) if precise, but 1.55 has 3 sig figs, so \( \frac{1550}{55.85} \approx 27.7 \) when rounded to 3 sig figs? Wait, 1550 is 1.55×10³ (3 sig figs), 55.85 is ~55.85. So \( 1550 \div 55.85 \approx 27.75 \), which rounds to 27.8 mol (3 sig figs).

Step1: Analyze stoichiometry

Reaction: \( 2 \, \text{Al} + 6 \, \text{HCl}
ightarrow 2 \, \text{AlCl}_3 + 3 \, \text{H}_2 \). Molar ratio of \( \text{Al} \) to \( \text{H}_2 \) is \( 2:3 \).

Step2: Calculate moles of Al

Let \( x \) be moles of Al. \( \frac{2}{3} = \frac{x}{6 \, \text{mol H}_2} \). Solve for \( x \): \( x = \frac{2 \times 6}{3} = 4 \, \text{mol} \).

Step1: Use dilution formula

Dilution formula: \( M_1 V_1 = M_2 V_2 \), where \( M_1 = 9.00 \, \text{M} \), \( V_2 = 6.5 \, \text{L} \), \( M_2 = 1.00 \, \text{M} \).

Step2: Solve for \( V_1 \)

\( V_1 = \frac{M_2 V_2}{M_1} = \frac{1.00 \, \text{M} \times 6.5 \, \text{L}}{9.00 \, \text{M}} \approx 0.722 \, \text{L} \) (3 sig figs from 9.00, 6.5 has 2? Wait, 6.5 is 2 sig figs, 9.00 is 3, 1.00 is 3. So \( \frac{1.00 \times 6.5}{9.00} = \frac{6.5}{9.00} \approx 0.722 \, \text{L} \) (but 6.5 has 2 sig figs? Wait, 6.5 L is two sig figs, 1.00 is three, 9.00 is three. The least number of sig figs in multiplication/division is two (from 6.5). Wait, no: 6.5 is two, 1.00 is three, 9.00 is three. So \( 6.5 \) has two, so result should have two? Wait, \( 1.00 \times 6.5 = 6.50 \) (three sig figs), divided by 9.00 (three sig figs) gives \( 6.50 / 9.00 = 0.7222... \), which with two sig figs would be 0.72 L? Wait, no: 6.5 is two sig figs, so \( V_1 = \frac{1.00 \times 6.5}{9.00} = \frac{6.5}{9.00} \approx 0.722 \, \text{L} \). But 6.5 has two sig figs, so maybe 0.72 L? Wait, 6.5 is two, 1.00 is three, 9.00 is three. The rule is that the number of sig figs is determined by the least precise measurement, which is 6.5 (two sig figs). So \( 6.5 / 9.00 = 0.722... \), rounded to two sig figs is 0.72 L. But let's check: \( M_1 V_1 = M_2 V_2 \) → \( 9.00 \times V_1 = 1.00 \times 6.5 \) → \( V_1 = 6.5 / 9.00 ≈ 0.722 \, \text{L} \). If 6.5 is two sig figs, then 0.72 L (two sig figs) or 0.722 L (three, if 6.5 is considered as two decimal places? No, 6.5 is two sig figs. Wait, 6.5 L: the 6 and 5 are significant, so two sig figs. 1.00 M is three, 9.00 M is three. So the answer should have two sig figs? Wait, no: when multiplying/dividing, the result has the same number of sig figs as the least precise measurement. 6.5 has two, so the answer should have two. So \( 6.5 / 9.00 = 0.7222... \) → 0.72 L. But let's do exact calculation: \( 1.00 \times 6.5 = 6.5 \), \( 6.5 / 9.00 = 0.722222... \). So with three sig figs (from 9.00 and 1.00, since 6.5 is two, but maybe 6.5 is considered as two, but 1.00 and 9.00 are three, so maybe the 6.5 is two, so the answer is two sig figs: 0.72 L. Or maybe the problem expects three sig figs because 9.00 and 1.00 have three, and 6.5 has two, but in dilution, sometimes the volume is considered as exact? Wait, the problem says "6.5 L" (two sig figs) and "9.00 M" (three), "1.00 M" (three). So the answer should have two sig figs? Wait, no, 6.5 is two, so the answer is two. But let's check: \( V_1 = (1.00 \, \text{M} \times 6.5 \, \text{L}) / 9.00 \, \text{M} = 6.5 / 9.00 ≈ 0.722 \, \text{L} \). If we take 6.5 as two sig figs, it's 0.72 L. If we take 6.5 as two decimal places (but it's not, it's two sig figs), so the answer is \( 0.72 \, \text{L} \) (two sig figs) or \( 0.722 \, \text{L} \) (three, if 6.5 is considered as two sig figs but the calculation allows three because 1.00 and 9.00 have three). The problem says "correct significant figures", so 6.5 has two, 9.00 and 1.00 have three. The rule is that the number of sig figs is determined by the least number, which is two. So \( 0.72 \, \text{L} \). But let's see: 6.5 is two, so 6.5 / 9.00 = 0.7222... → 0.72 L (two sig figs).

Answer:

\( 27.8 \, \text{mol} \) (or \( 27.7 \, \text{mol} \) depending on rounding; precise calculation: \( \frac{1550}{55.845} \approx 27.75 \), so 27.8 mol with 3 sig figs)

Question 41 (Moles of Aluminum):