Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete

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Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete

combustion of butane is 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 1.40 g of butane?

Chemistry

2 answers:

zavuch27 [327] · Answer 1 · 2022-05-12T00:55:41+03:00

Answer:

So the volume will be 2.33 L

Explanation:

The reaction for the combustion is:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)

mass of butane to moles (mass / molar mass)

1.4 g / 58 g/mol

= 0.024 moles

2 moles of butane can produce 8 moles of carbon dioxide

0.024 moles of butane must produce (0.024 × 8) /2

= 0.096 moles of CO₂

Now we apply the Ideal Gases Law to find out the volume formed.

P . V = n . R . T

p = 1atm

n = 0.096 mol

R = 0.082 L.atm/mol.K

T = 273 + 23 = 296K

V = ?

1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K

V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm

= 2.33 L

So the volume will be 2.33 L

Sever21 [200] · Answer 2 · 2022-05-12T02:45:57+03:00

Answer:

the volume of carbon dioxide formed by the combustion of 1.40 g of butane is 2.3447 L

Explanation:

Given:

2C₄H₁₀(g) + 13O₂(g) ⇒ 8CO₂(g) + 10H₂O(l)

pressure(P) = 1.00 atm

temperature(t) = 23°C

Mass of butane(m) = 1.40 g

From the balanced equation of the complete combustion of butane(C₄H₁₀), 2 moles of butane produce 8 moles of carbon dioxide(CO₂). Therefore 1 mole of butane would produce 2 mole of carbon dioxide

Therefore The molar mass of butane(M) = M(C₄H₁₀) = (12 ×4) + (1 × 10) = 48 + 10 = 58 g/mol

The amount of mole of butane(n) = mass of butane(m)/molar mass of butane(M)

n(C₄H₁₀) = m/M

n(C₄H₁₀) = 1.4/58 = 0.0241 mole

since 1 mole of butane gives 4 mole of carbon dioxide, therefore 0.0241 mole of butane = 4 × 0.0241 mole of carbon dioxide.

n(CO₂) = 4 × n(C₄H₁₀) = 4 × 0.0241 = 0.0966 mole

volume of carbon dioxide formed by the combustion of 1.40 g of butane (V) is given by

PV = nRT

∴ V = nRT/P

T = 23°C = 23 + 273 = 296 K

R = 0.08205 L atm mol⁻¹ K⁻¹

since V = nRT/P