Answer: 3.09 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2SO_4=\frac{2.70g}{98g/mol}=0.0275moles[/tex]
The balanced chemical equation is
[tex]2KOH+H_2SO_4\rightarrow 6H_2O+K_2SO_4[/tex]
According to stoichiometry :
1 mole of [tex]H_2SO_4[/tex] require 2 moles of [tex]KOH[/tex]
Thus 0.0275 moles of [tex]H_2SO_4[/tex] will require=[tex]\frac{2}{1}\times 0.0275=0.0551moles[/tex] of [tex]KOH[/tex]
Mass of [tex]KOH=moles\times {\text {Molar mass}}=0.0551moles\times 56g/mol=3.09g[/tex]
Thus 3.09 g of KOH is required to react completely with 2.70 grams of
[tex]H_2SO_4[/tex]
