Question

H₂ (2)
→ H.
H₂(8)
+
2(g)
wer

Answer 1

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
`H_2`

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} H_2=(8g)/(2g/mol)=4moles`

`\text{Moles of} O_2=(16g)/(32g/mol)=0.5moles`

`2H_2(g)+O_2(g)\rightarrow 2H_2O(g)`

According to stoichiometry :

1 moles of
`O_2` require 2 moles of
`H_2`

Thus 0.5 moles of
`O_2` will require=
`(2)/(1)* 0.5=1.0moles` of
`H_2`

Thus
`O_2` is the limiting reagent as it limits the formation of product and
`H_2` is the excess reagent.

(4.0-1.0) = 3.0 moles of are left unreacted

Mass of remained=

Thus 6.0 g of
`H_2` will remain.