Respuesta :
Answer: 86.2 g of [tex]BaSO_4[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]Ba(OH)_2[/tex]
[tex]\text{Number of moles}=\frac{67.1g}{171g/mol}=0.39moles[/tex]
b) moles of [tex]H_2SO_4[/tex]
[tex]\text{Number of moles}=\frac{36.3g}{98g/mol}=0.37moles[/tex]
[tex]Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O[/tex]
According to stoichiometry :
1 moles of [tex]H_2SO_4[/tex] require 1 mole of [tex]Ba(OH)_2[/tex]
Thus 0.37 moles of [tex]H_2SO_4[/tex] require=[tex]\frac{1}{1}\times 0.37=0.37moles[/tex] of [tex]Ba(OH)_2[/tex]
Thus [tex]H_2SO_4[/tex] is the limiting reagent as it limits the formation of product.
As 1 mole of [tex]H_2SO_4[/tex] give = 1 mole of [texBaSO_4[/tex]
Thus 0.37 moles of [tex]H_2SO_4[/tex] give =[tex]\frac{1}{1}\times 0.37=0.37moles[/tex] of [tex]BaSO_4[/tex]
Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.37moles\times 233g/mol=86.2g[/tex]
Thus 86.2 g of [tex]BaSO_4[/tex] will be produced from the given masses of both reactants.
