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?
Answers
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.