Question

80 g of H2 is reacted with 80g of O2 to form water. Find out the mass of water obtained. Which substance is the limiting reagent

Answer 1

H2 is 2 gram
so 40H2

O2 is 32 gram
so 2.5O2

H20 is water
maximum is 2.5H2O
weight of 2.5H2O is 45

o2 is limiting reagent

Answer 2

Answer: oxygen

Explanation:

$2H_2+O_2\rightarrow 2H_2O$

Moles is calculated by using the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $H_2$ = 80 g

Molar mass of $H_2$ = 2 g/mol

$\text{Moles of hydrogen}=\frac{80g}{2g/mol}=40moles$

Given mass of $O_2$ = 80 g

Molar mass of $O_2$ = 32 g/mol

$\text{Moles of hydrogen}=\frac{80g}{32g/mol}=2.5moles$

From the given balanced equation it is seen that 2 moles of oxygen is reacted with 1 mole of hydrogen

2.5 moles of oxygen will react with=$\frac{1}{2}\times 2.5=5mole$ of hydrogen

Thus oxygen is the limiting reagent as it limits the formation of product. Hydrogen is in excess as (40-5) =35 moles are left as such.