1) First let's write the equation. It is a combustion reaction, so:
C₃H₈ + O₂ ---> CO₂ + H₂O
and balance the equation (same number of atoms of each element on both sides of the equation):
C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O
Reactant side:
C - 3
H - 8
O - 10
Product side:
C - 3
O - 10
H - 8
2) Now let's transform 85 grams of CO₂ into mole. For this, we use the following equation:
mole = mass/molar mass
Molar mass of CO₂ is: (1×12) + (2×16) = 44 g/mol
mole = 85/44
mole = 1.9 mol of CO₂
3) Now we use the proportion of the balanced equation:
1 mol of C₃H₈ ---- 3 mol of CO₂
x mol of C₃H₈ ----- 1.9 mol of CO₂
x = 0.6 mol of C₃H₈
4) Now we transform mole of C₃H₈ into grams using its molar mass.
molar mass of C₃H₈ is: (3×12) + (8×1) = 44 g/mol
mass = mole × molar mass
mass = 0.6 × 44
mass of C₃H₈ = 28 g
Answer: a) mass of C₃H₈ = 28 g
For alternative b we follow the same process starting from step 3:
3)Now we use the proportion of the balanced equation:
5 mol of O₂ ---- 3 mol of CO₂
x mol of O₂ ----- 1.9 mol of CO₂
x = 3.16 mol of O₂
4) Now we transform mole of O₂ into grams using its molar mass.
molar mass of O₂ is: (2×16) = 32 g/mol
mass = mole × molar mass
mass = 3.16 × 32
mass of O₂ = 101 g
Answer: b) mass of O₂ = 101 g
For alternative c we follow the same process starting from step 3:
3) Now we use the proportion of the balanced equation:
4 mol of H₂O ---- 3 mol of CO₂
x mol of H₂O ----- 1.9 mol of CO₂
x = 0.84 mol of H₂O
4) Now we transform mole of H₂O into grams using its molar mass.
molar mass of H₂O is: (2×1) + (1×16) = 18 g/mol
mass = mole × molar mass
mass = 0.84 × 18
mass of H₂O = 15 g
Answer: c) mass of H₂O = 15 g