Question

Hydrogen and oxygen combine in the ratio of 1:8 by mass to from water what mass of oxygen gas would be required to react completely with 3g of hydrogen gas?

Answer 1

Explanation:

: The weight of excess reactant left is, (B) 1.0 g of oxygen.

Solution : Given,

Mass of Mg = 3 g

Mass of O_2O

2

= 3 g

Molar mass of Mg = 24 g/mole

Molar mass of O_2O

2

= 32 g/mole

First we have to calculate the moles of MgMg and O_2O

2

\text{ Moles of Mg}=\frac{\text{ Mass of Mg}}{\text{ Molar mass of Mg}}=\frac{3g}{24g/mole}=0.125moles Moles of Mg=

Molar mass of Mg

Mass of Mg

=

24g/mole

3g

=0.125moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.094moles Moles of O

2

=

Molar mass of O

2

Mass of O

2

=

32g/mole

3g

=0.094moles

The balanced chemical reaction is,

2Mg+O_2\rightarrow 2MgO2Mg+O

2

→2MgO

From the reaction, we conclude that

2 moles of MgMg react with 1 moles of O_2O

2

0.125 moles of MgMg react with \frac{0.125}{2}=0.0625

2

0.125

=0.0625 moles of O_2O

2

So, oxygen is present in excess amount and magnesium in less amount.

In this reaction, MgMg is limiting reactant and O_2O

2

is excess reactant.

Moles of excess reactant left = 0.094 - 0.0625 = 0.0315 moles

Weight of excess reactant left = Moles of excess reactant left × Molar mass of excess reactant

Weight of excess reactant left = 0.0315 g × 32 g/mole = 1.008 g = 1.0 g

Therefore, the weight of excess reactant left is, (B) 1.0 g of oxygen.