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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.
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