Question

Determine the mass in grams of CO₂ that is produced by the complete reaction of 0.08142 moles of C₅H₁₂ (pentane) according to the following combustion reaction:

C₅H₁₂(l) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g)

Answer 1

Answer: 17.912 gm

Explanation: C5H12 + 8O2 ------- 5CO2 +6H2O

IF ONE MOLE OF C5H12 produces 5 moles of CO2 then 0.08142 produces x mole

then x = 5 X 0.08142

X = 0.4071

NO. OF MOLES = given mass/ molecular mass of CO2

0.4071= mass/44

Mass= 0.4071 X 44

mass = 17.912

therefore mass of CO2 produced is 17.912

Answer 2

Answer: 17.91 grams of CO2 will be produced

Step-by-step explanation:

The balanced chemical equation for the combustion of pentane is:

C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)

From the equation, we can see that for every 1 mole of C5H12 that reacts, 5 moles of CO2 are produced. Therefore, to determine the mass of CO2 produced, we need to first calculate the number of moles of CO2 produced by 0.08142 moles of C5H12:

0.08142 moles C5H12 x (5 moles CO2 / 1 mole C5H12) = 0.4071 moles CO2

Now we can use the molar mass of CO2 (44.01 g/mol) to calculate the mass of CO2 produced:

0.4071 moles CO2 x 44.01 g/mol = 17.91 g CO2

Therefore, the mass of CO2 produced by the complete combustion of 0.08142 moles of pentane is 17.91 g.