Respuesta :
Answer: The limiting reactant is hydrogen gas and the mass of excess reactant [tex](SnO_2)[/tex] left over is 197.43 g
Explanation:
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(1)
We are given:
Given mass of [tex]SnO_2[/tex] = 351.3 g
Molar mass of [tex]SnO_2[/tex] = 150.71 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }SnO_2=\frac{351.3g}{150.71g/mol}=2.33mol[/tex]
At STP conditions:
22.4 L of volume is occupied by 1 mole of a gas
Applying unitary method:
45.8 L of volume will be occupied by = [tex]\frac{1mol}{22.4L}\times 45.8L=2.04mol[/tex] of hydrogen gas
For the given chemical reaction:
[tex]SnO_2+2H_2\rightarrow Sn+2H_2O[/tex]
By stoichiometry of the reaction:
If 2 moles of hydrogen gas reacts with 1 mole of [tex]SnO_2[/tex]
So, 2.04 moles of hydrogen gas will react with = [tex]\frac{1}{2}\times 2.04=1.02mol[/tex] of [tex]SnO_2[/tex]
As the given amount of [tex]SnO_2[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.
Moles of excess reactant ([tex]SnO_2[/tex]) left = [2.33 - 1.02] = 1.31 moles
We know, molar mass of [tex]SnO_2[/tex] = 150.71 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }SnO_2=(1.31mol\times 150.71g/mol)=197.43g[/tex]
Hence, the limiting reactant is hydrogen gas and the mass of excess reactant [tex](SnO_2)[/tex] left over is 197.43 g
