Question

How many grams of 02 are needed to completely burn 26.2 g of C3H8

Answer 1

Molar mass :

O2 = 31.99 g/mol

C3H8 = 44.09 g/mol

C3H8 + 5 O2 = 3 CO2 + 4 H2O

44.09 g ---------------- 5 x 31.99 g
26.2 g ----------------- ?? ( mass of O2)

26.2 x 5 x 31.99 / 44.09 =

4190.69 / 44.09 => 95.04 g of O2

hope this helps!

Answer 2

Answer:

We need 95.0 grams of O2

Explanation:

Step 1 :Data given

Mass of C3H8 = 26.2 grams

Molar mass C3H8  = 44.1 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 3: Calculate moles C3H8

Moles propane = mass / molar mass

Moles propane = 26.2 grams / 44.1 g/mol

Moles propane = 0.594 moles

Step 4: Calculate moles O2

For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 0.594 moles propane we need 5*0.594 = 2.97 moles O2

Step 5: Calculate mass O2

Mass O2 = moles * molar mass

Mass O2 = 2.97 moles * 32.0 g/mol

Mass O2 = 95.0 grams

We need 95.0 grams of O2