Question

strontium oxide reacts with liquid water to produce strontium hydroxide.
sro$_{(s)}$ + h$_2$o$_{(l)}$ $\
ightarrow$ sr(oh)$_2

$$_{(aq)}$ when 1.701 g of sro is added to 48.96 ml of water, the temperature of the solution rose from 20.21$\\degree$c to 30.92$\\degree$c in a coffee cup calorimeter. assuming the density of water at this temperature is 1.000 g/ml, the specific heat of the solution is 3.97 j/($\\degree$c·g) and the heat absorbed by the constant pressure calorimeter is negligible, calculate the molar enthalpy change for this reaction. report your answer in kj/mol to 3 sig figs, but do not include units in your answer. question 2 0.5 pts a sample of acetone, ch$_3$coch$_3$, weighing 2.048 g was burned in excess oxygen in a bomb calorimeter. given that the temperature of the calorimeter rose from 20.03$\\degree$c to 24.74$\\degree$c and the heat capacity of the calorimeter and its contents was 13.05 kj/$\\degree$c, what is the heat produced per mole of acetone? the combustion of acetone is given by: ch$_3$coch$_3$$

_{(l)}$ + 4 o$_2

$$_{(g)}$ $\ ightarrow$ 3 co$_2$$

_{(g)}$ + 3 h$_2$o$_{(l)}$
report your answer in kj/mol to 4 sig figs, but do not include units in the answer.

Explanation:

Question 1 (Strontium Oxide Reaction)

Step1: Calculate mass of solution

Mass of water = volume × density = \( 48.96\space mL × 1.000\space g/mL = 48.96\space g \)
Mass of \( SrO = 1.701\space g \)
Total mass of solution (\( m \)) = \( 48.96 + 1.701 = 50.661\space g \)

Step2: Calculate heat absorbed by solution (\( q \))

Using \( q = mc\Delta T \), where \( c = 3.97\space J/(\degree C·g) \), \( \Delta T = 30.92 - 20.21 = 10.71\space \degree C \)
\( q = 50.661\space g × 3.97\space J/(\degree C·g) × 10.71\space \degree C \)
\( q = 50.661×3.97×10.71 ≈ 2170.5\space J = 2.1705\space kJ \) (exothermic, so \( q_{reaction} = -q_{solution} \))

Step3: Calculate moles of \( SrO \)

Molar mass of \( SrO = 87.62 + 16.00 = 103.62\space g/mol \)
Moles (\( n \)) = \( \frac{1.701\space g}{103.62\space g/mol} ≈ 0.01642\space mol \)

Step4: Calculate molar enthalpy change (\( \Delta H \))

\( \Delta H = \frac{q_{reaction}}{n} = \frac{-2.1705\space kJ}{0.01642\space mol} ≈ -132 \)

Question 2 (Acetone Combustion)

Step1: Calculate heat absorbed by calorimeter (\( q \))

Using \( q = C\Delta T \), where \( C = 13.05\space kJ/\degree C \), \( \Delta T = 24.74 - 20.03 = 4.71\space \degree C \)
\( q = 13.05\space kJ/\degree C × 4.71\space \degree C ≈ 61.4655\space kJ \) (exothermic, \( q_{reaction} = -61.4655\space kJ \))

Step2: Calculate moles of acetone (\( CH_3COCH_3 \))

Molar mass of acetone = \( 3×12.01 + 6×1.008 + 16.00 = 58.08\space g/mol \)
Moles (\( n \)) = \( \frac{2.048\space g}{58.08\space g/mol} ≈ 0.03526\space mol \)

Step3: Calculate heat per mole

\( \Delta H = \frac{q_{reaction}}{n} = \frac{-61.4655\space kJ}{0.03526\space mol} ≈ -1743 \)

Answer:

s:
Question 1: \(-132\)
Question 2: \(-1743\)