Question

1. How many milliliters of O2 are consumed in the complete combustion of a sample of hexane, C6H14, if the reaction produces 516 mL of CO2? Assume all gas volumes are measured at the same temperature and pressure. The reaction is: 2C6H14(g) + 19O2(g) ----> 12CO2(g) + 14H2O(g)

2. How many milliliters of O2 measured at 27.0 °C and 654 torr are needed to react completely with 16.8 mL of CH4 measured at 35.0 °C and 725 torr?

3. If 16.0 g of water is converted to steam in a 4.60 L pressure cooker held at a temperature of 115 °C, what pressure (in atm) would be produced?

Answer 1

1)

2C6H14(g) + 19O2(g) ----> 12CO2(g) + 14H2O(g)

19 moles of O2 will produce 12 moles of CO2

Volume of O2 consumed = Volume of CO2 * 19/12 = 516 * 19/12 = 817 ml

2)

Using ideal gas equation for CH4

(725/760) * 16.8 = n * 0.0821 * (273.15+35)

n = 6.3347 * 10^(-4) moles

Moles of O2 = 2 * 6.3347 * 10^(-4) moles = 1.266 * 10^(-3) moles

(654/760) * Volume = 1.266 * 10^(-3) moles * 0.0821 * 300.15

Volume = 36.253 mL

3) Using ideal gas equation

PV = nRT

Molar mass of water = 18 gm/mol

Number of moles of water = 16/18 = 0.88 moles

P(4.60) = 0.88 * 0.0821 * (273.15+115)

P = 6.096 atm