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?

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

substituting values,

V = (0.0966 × 0.0820 × 296)/1 = 2.3447

V = 2.3447 L

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates

2b2t

explanation:

2b2t