The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol. How many grams of CO2 are produced per 1.00 × 104 kJ of heat released?

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:

15.1 mol × 44.01 g/mol = 665 g

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer