chem homework help

Question

A sample of 6.35 g6.35 g of liquid 1‑propanol, C3H8O,C3H8O, is combusted with 43.7 g43.7 g of oxygen gas. Carbon dioxide and water are the products.

Write the balanced chemical equation for the reaction. Physical states are optional.

What is the limiting reactant?

How many grams of CO2 are released in the reaction?

How many grams of the excess reactant remain after the reaction is complete?

Answer 1

C₃H₈O + 5O₂ -> 3CO₂ + 4H₂O

Calculate Mole Ratio: moles / (coefficient of balanced equation):

mole ratio of C₃H₈O -> 6.35/(36+8+16)/1 = 0.106

mole ratio of O₂ -> 43.7/32/5 = 0.237

C₃H₈O -> Limiting reagent

Use 0.106 moles of C₃H₈O to determine moles of CO₂. 3 more moles of CO2 for 1 mole of C₃H₈O , so multiply by 3:

moles of CO₂ = 0.106* 3 = 0.318 moles

Multiply by MW to get grams:

0.318*(44) = 13.99 grams

Find moles of O₂ for 0.106 moles of C₃H₈O . 5 moles of O₂ for every mole of C₃H₈O based on balanced equation, so multiply by 5:

moles of O₂ = 0.106*5 = 0.53 moles

grams of O₂ from above moles = 0.53*32g = 16.96g

Subtract 43.7 by 16.96 to get excess = 26.74g