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Determine the mass of carbon dioxide produced when 0.85 grams of butane (C4H10)reacts with oxygen according to the following balanced chemical equation: 2 C4H10 (1) + 13 O2 (g) --> 8 CO2 (g) + 10 H2O (g)

2.58g
4.79g
0.03g
15.75g

Sagot :

Victoronyangooketchidn

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]

What are moles?

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.

Learn more about the moles here:

https://brainly.com/question/15209553

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