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Answer:
2.58g
Explanation:
First calculate the moles of butane used in the reaction
moles = mass÷molar mass
= 0.85÷58
= 0.0147
According to the stoichiometric ratio:
C4H10 : CO2 = 2:8
moles of CO2 =(8÷2)×0.0147
=0.0586 moles
mass of CO2 = 0.0586×44
= 2.58g
2.58g is the mass of carbon dioxide produced when 0.85 grams of butane ([tex]C_4H_{10}[/tex]) reacts with oxygen according to the following balanced chemical equation: [tex]2C_4H_{10} + 13O_2 - > 8CO_2 (g) + 10H_2O (g)[/tex]
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
Firstly calculate the moles of butane used in the reaction
moles = [tex]\frac{mass}{molar \;mass}[/tex]
= [tex]\frac{0.85}{58}[/tex]
= 0.0147
According to the stoichiometric ratio:
[tex]C_4H_{10}[/tex] : [tex]CO_2[/tex] = 2:8
moles of [tex]CO_2[/tex] = ([tex]\frac{8}{2}[/tex]) × 0.0147
=0.0586 moles
mass of [tex]CO_2[/tex] = 0.0586×44
= 2.58g
Hence, option A is correct.
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