O 2 and C 3 H 8 react according to the following unbalanced chemical equation: O

Question

O2 and C3H8 react according to the following unbalanced chemical equation:

O2 + C3H8 CO2 + H2O

Given 8.27 g of O2 and 8.32 g of C3H8, answer the following questions: (For the calculations, provide answers with 3 significant figures.)


1. Balance the chemical equation using the lowest integer coefficients.

5 O2 + 1 C3H8   3 CO2 + 4 H2O correct

2. Calculate the mass (in g) of the excess reagent remaining after the reaction is complete

3. Calculate the theoretical yield of H2O (in g) from the reaction

4. Determine the percent yield of H2O if 2.24 g of H2O is actually produced from the reaction.

Explanation / Answer

5 O2 + 1 C3H8   3 CO2 + 4 H2O

2. First, calculate the limitant reactant.

Moles of O2 = 8.27 g /32 g = 0.2584 mol O2

moles C3H8 = 8.32g/44g = 0.189 mol C3H8.

0.2584/0.189 = 1.36

and 5/1 = 5.

1.36 < 5. Limitant reactive is O2

8.32 g O2 * (1mol/32g)*(1 mol C3H8/5 mol O2)*(44g/imol C3H8) = 2.28 g O2

8.32 g O2 * (1mol O2/32g)*(4 molH2O/5molO2)*(18 g /1 molH2O) = 3.77 g H2O

2.24/3.77 * 100 = 59.4 %

8.32 g O2 * (1mol/18g)

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.