IDNLearn.com is designed to help you find reliable answers to any question you have. Find in-depth and trustworthy answers to all your questions from our experienced community members.
Sagot :
The given question is incomplete. The complete question is:
In the chemical reaction: , with 8 grams of and 16 grams of and the reaction goes to completion, what is the excess reactant and how much of that would remain?
A) 6 grams of
B) 7 grams of
C) 8 grams of
D) 12 grams of
E) 14 grams of
Answer: A) 6 grams of [tex]H_2[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{8g}{2g/mol}=4moles[/tex]
[tex]\text{Moles of} O_2=\frac{16g}{32g/mol}=0.5moles[/tex]
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]
According to stoichiometry :
1 moles of [tex]O_2[/tex] require 2 moles of [tex]H_2[/tex]
Thus 0.5 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.5=1.0moles[/tex] of [tex]H_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
(4.0-1.0) = 3.0 moles of are left unreacted
Mass of remained=
Thus 6.0 g of [tex]H_2[/tex] will remain.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
