Question

The combustion of propane can be represented by the following equation: __ C3H8 (g) + __O2(g) → __CO2(g) + __H2O(g) Balance the equation and answer the next three questions. If 257. g of C3H8 and 951. g O2 react, which reactant is used up first? (enter propane or oxygen) What is the mass of carbon dioxide produced? (Give your answer to 3 sig figs.) What is the mass of water produced?(Give your answer to 3 sig figs.) grams

Answer 1

C3H8(g) + O2(g) ----------------- CO2(g) + H2O(g)

C3H8(g) + 5 O2(g) ----------------- 3 CO2(g) + 4 H2O(g)   balanced equation

mass of C3H8= 257.0 grams

molar mass of C3H8 = 44.1gram/mole

number of moles of C3H8 = mass/molar mass = 257.0/44.1= 5.828 moles

mass of O2= 951.0 grams

molar mass of O2= 32.0 gram/mole

number of moles of O2= 951.0/32.0=29.72 moles

according to equation

1 mole of C3H8 = 5 moles of O2

5.828 moles of C3H8 = ?

                                  = 5.828 x 5/1 = 29.14 moles of O2.

we need 29.14 moles of O2. But we have 29.72 moles of O2.SO is excess reagent.

hence C3H8 is limiting reagent

Hence C3H8 is used up first in the reaction.

b)

according to equation

1 mole of C3H8 = 3 moles of CO2

5.828 moles of C3H8 = ?

                                  = 5.828 x 3/1 = 17.484 moles of CO2

number of moles of CO2 formed = 17.484 moles

molar mass of CO2 = 44.1 gram/mole

mass of one mole of CO2 = 44.1 grams

mass of 17.484 moles of CO2 = ?

                                                 = 17.484 x 44.1= 771.0 grams

mass of CO2 produced = 771 grams

c)

according to equation

1 mole of C3H8 = 4 moles of H2O

5.828 moles of C3H8 = ?

                                          = 5.828 x 4 = 23.312 moles of H2O

number of moles of H2O formed = 23.312 moles

molar mass of H2O = 18.0gram

mass of 23.312 moles of H2O = 23.312 x 18.0=419.6 grams

mass of H2O formed = 420 grams.