Calculate the mass of butane needed toproduce 82.1 g of carbondioxide. Solution

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HERE IS A SIMILAR QUESTION: 2C4H10+13O2=10H2O+8CO2 Calculate the mass of butane needed to produce 45.1 g of carbondioxide. 2C4H10 + 13O2 ---> 10H2O + 8CO2 if n(X), m(X) and M(X) are the moles, mass and molar mass ofspecies X then n(C4H10) = m(C4H10) / M(C4H10) ==> m(C4H10) = n(C4H10)M(C4H10) for every mole of butane burned, you get 4 moles of CO2, i.e. n(CO2) = 4n(C4H10) and n(CO2) = m(CO2) / M(CO2) ==> m(C4H10) =m(CO2)M(C4H10) /4M(CO2) m(CO2) = 45.1 g M(C4H10) = 58.12 g mol¯¹ M(CO2) = 44.01 g mol¯¹ Substituting gives: m(C4H10) = 14.89 g Mass of butane needed is 14.9 g (to 3 s.f.).

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