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 and 735 mmHg? [Hint: What is the standard molar enthalpy of combustion of C3H8(g)?]

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)?]

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:

where,

P = pressure of the gas = 735 mmHg

V = Volume of the gas = 1.60 L

T = Temperature of the gas =

R = Gas constant =

n = number of moles of propane = ?

Putting values in above equation, we get:

The equation used to calculate enthalpy change is of a reaction is:

The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

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 =

To calculate the moles of ice, we use the equation:

where,

= amount of heat released = 142.07 kJ

n = number of moles of ice = ?

= molar heat of fusion = 6.01 kJ/mol

Putting values in above equation, we get:

To calculate the number of moles, we use the equation:

Molar mass of ice = 18 g/mol

Moles of ice = 23.64 moles

Putting values in above equation, we get:

Hence, the mass of ice that would be melted is 425.52 grams

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