Answer:
Mass of butane = 1.87 g
Explanation:
Form a balanced chemical equation for the reaction stated:
C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O
Number of moles of CO₂
[tex] = \frac{mass}{molar \: mass} [/tex]
[tex] = \frac{15}{12 + 2(16)} [/tex]
[tex] = 0.34 \: mol[/tex]
From the equation,
1 mol C₄H₁₀ : 4 mol CO₂
0.085 mol C₄H₁₀ : 0.34 mol CO₂
Mass of butane
[tex] = number \: of \: moles \times molar \: mass[/tex]
[tex] = 0.085 \times (12 + 10(1))[/tex]
[tex] = 1.87 \: g[/tex]
The grams of butane must be burned in an excess of Oxygen, to produce 15g of CO₂ is 1.87 grams.
Relation between mass and moles will be represented as:
n = W/M, where
W = require or given mass
M = molar mass
Given balance chemical reaction is:
C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O
Moles of 15g of CO₂ = 15g / 44g/mol = 0.34 moles
From the stoichiometry of the reaction, it is clear that:
4 moles of CO₂ = produced by 1 mole of C₄H₁₀
0.34 moles of CO₂ = produced by 1/4×0.34=0.085 moles of C₄H₁₀
Now we convert moles of butane to mass as:
W = (0.085)(22) = 1.87 g
Hence required mass of butane is 1.87 grams.
To know more about mass & moles, visit the below link:
https://brainly.com/question/24639749
