How many grams of \text{CO}_2CO 2 start text, C, O, end text, start subscript, 2, end subscript will be produced from 39.0 \text{ g}39.0 g39, point, 0, start text, space, g, end text of \text{C}_3\text{H}_8C 3 H 8 start text, C, end text, start subscript, 3, end subscript, start text, H, end text, start subscript, 8, end subscript and 11.0 \text{ g}11.0 g11, point, 0, start text, space, g, end text of \text{O}_2O 2

7.568g of CO₂ will be produced

Explanation:

Based on the reaction:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Where 1 mole of  C₃H₈ react with 5 moles of O₂ to produce 3 moles of CO₂ and 4 moles of H₂O

Moles of C₃H₈ and O₂ are:

C₃H₈: 39.0g ₓ (1mol / 44.1g) = 0.884 moles

O₂: 11.0g ₓ (1mol / 32g) = 0.344 moles

For a complete reaction of 0.884 moles of C₃H₈ are necessaries:

0.884 mol C₃H₈ ₓ (6 mol O₂ / 1 mol C₃H₈) = 5.304 moles of O₂

As you have just 0.344 moles of O₂, it is the limiting reactant.

And moles of CO₂ produced are:

0.344 mol O₂ ₓ (3 mol CO₂ / 6 mol O₂) = 0.172 moles of CO₂

Thus, mass in grams of CO₂ (44g / mol) is:

0.172 moles of CO₂ ₓ (44g / mol) = 7.568g of CO₂ will be produced