Answer :
The question is incomplete, here is the complete question:
An ice calorimeter measures quantities of heat by the quantity of ice melted. How many grams of ice would be melted by the heat released in the complete combustion of 1.60 L of propane gas, measured at 20.0 °C and 735 mmHg? [Hint: What is the standard molar enthalpy of combustion of C₃H₈(g)?]
Answer: The mass of ice that would be melted is 425.52 grams
Explanation:
- To calculate the moles of propane, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 735 mmHg
V = Volume of the gas = 1.60 L
T = Temperature of the gas = [tex]20^oC=[20+273]K=293K[/tex]
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of propane = ?
Putting values in above equation, we get:
[tex]735mmHg\times 1.60L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735\times 1.60}{62.3637\times 293}=0.064mol[/tex]
- The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]
The chemical equation for the combustion of propane follows:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(3\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_3H_8(g))})+(5\times \Delta H_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(C_3H_8(g))}=-103.8kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[(3\times (-393.5))+(4\times (-285.8))]-[(1\times (-103.8))+(5\times (0))]\\\\\Delta H_{rxn}=-2219.9kJ[/tex]
By Stoichiometry of the reaction:
When 1 mole of propane is combusted, the heat released is 2219.9 kJ
So, when 0.064 moles of propane is combusted, the heat released will be = [tex]\frac{2219.9}{1}\times 0.064=142.07kJ[/tex]
- To calculate the moles of ice, we use the equation:
[tex]\Delta H_{fusion}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released = 142.07 kJ
n = number of moles of ice = ?
[tex]\Delta H_{fusion}[/tex] = molar heat of fusion = 6.01 kJ/mol
Putting values in above equation, we get:
[tex]6.01kJ/mol=\frac{142.07kJ}{n}\\\\n=\frac{142.07kJ}{6.01kJ/mol}=23.64mol[/tex]
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of ice = 18 g/mol
Moles of ice = 23.64 moles
Putting values in above equation, we get:
[tex]23.64mol=\frac{\text{Mass of ice}}{18g/mol}\\\\\text{Mass of ice}=(23.64mol\times 18g/mol)=425.52g[/tex]
Hence, the mass of ice that would be melted is 425.52 grams