6. The combustion of a sample of butane, C. H. (lighter fluid), produced 2.46 grams of water. Using the balanced equation below, calculate the mass (grams) of carbon
dioxide produced.
2 C. H. + 13 0, > 8 CO, +10 H2O.

please help me find the carbon dioxide :)

Hey,

QUESTION)

1) We balance the equation of the reaction, that to say :

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

2) Calculating the molar mass of simple butane :

M(C₄H₁₀) = 4M(C) + 10M(H)

M(C₄H₁₀) = 4 x 12,0 + 10,0 M(C₄H₁₀) = 58,0 g/mol

3) Let us calculate the quantity of material corresponding to the mass of simple butane :

n(C₄H₁₀) = m/M

n(C₄H₁₀) = 2.46/58,0 n(C₄H₁₀) = 4,24 x 10⁻² mol

4) Determining the progress maximal x of the reaction,

xmax = n(C₄H₁₀)/2

xmax = 4,24 x 10⁻²/2 xmax = 2,12 x 10⁻² mol

5) If the reagents are supplied in a stoichiometric quantity, and the combustion is complete, then :

n(CO₂) = xmax x 8

n(CO₂) = 2,12 x 10⁻² x 8 n(CO₂) = 17,0 x 10⁻² mol

m(CO₂) = M(CO₂) x n(CO₂)

m(CO₂) = (12 + 2 x 16) x 17,0 x 10⁻² m(CO₂) = 7,48 g The mass of carbon dioxide is therefore 7.48 g.

The answer is b metal denting

Answerδh > 0explanationc₆h₆  (s)  → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.