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C7H16 + 11O2 = 7CO2 + 8H2O

Suppose 0.470kg of heptane are burned in air at a pressure of exactly 1atm and a temperature of 17.0°C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

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Step-by-step explanation:

First, we need to calculate the number of moles of heptane that are burned:

0.470 kg heptane × (1 mol heptane / 100.2057 g heptane) = 0.004696 mol heptane

According to the balanced chemical equation, 1 mol of heptane produces 7 mol of CO2. Therefore, the number of moles of CO2 produced is:

0.004696 mol heptane × (7 mol CO2 / 1 mol heptane) = 0.032872 mol CO2

The ideal gas law relates the number of moles of a gas to its volume, pressure, and temperature:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We can rearrange the ideal gas law to solve for the volume:

V = nRT / P

where V is in liters, n is in moles, R is 0.08206 L·atm/mol·K (the ideal gas constant), T is in Kelvin, and P is in atm.

We need to convert the temperature to Kelvin:

17.0°C + 273.15 = 290.15 K

Substituting the values into the equation, we get:

V = (0.032872 mol CO2)(0.08206 L·atm/mol·K)(290.15 K) / 1 atm = 0.758 L

Therefore, the volume of carbon dioxide gas produced is 0.758 L. Rounded to 3 significant digits, the answer is 0.758 L.

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