Question #3870a

Start by writing the balanced chemcial equation for the combustion of propane, #C_3H_8#

#C_3H_(8(g)) + 5O_(2(g)) -> color(red)(3)CO_(2(g)) + 4H_2O_((g))#

Notice that you have a #1:color(red)(3)# mole ratio between propane and carbon dioxide. This means that, regardless of how many moles of propane react, the reaction will always produce 3 times more moles of carbon dioxide.

Assuming that the oxygen is in excess, you can determine how many moles of propane react by using propane's molar mass

#42cancel("kg") * (1000cancel("g"))/(1cancel("kg")) * "1 mole"/(44.1cancel("g")) = "952.4 moles"# #C_3H_8#

This means that the reaction will produce

#952.4cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole"C_3H_8)) = "2857.2 moles"# #CO_2#

To determine the mass of the produced carbon dioxide, use carbon dioxide's molar mass

#2857.2cancel("moles") * "44.01 g"/(1cancel("mole")) = "125,745.4 g"# #CO_2#

Rounded to two sig figs, the number of sig figs you gave for the mass of propane, and expressed in kilograms, the answer will be

#125,754.4cancel("g") * "1 kg"/(1000cancel("g")) = color(green)("130 kg")#