Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Ask your questions and receive precise answers from experienced professionals across different disciplines. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Help??

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 :

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

#SPJ2

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.