Answer: 15.8 g of [tex]KMnO_4[/tex] will be required to produce 1.60 grams of [tex]O_2[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles[/tex]
[tex]2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2[/tex]
According to stoichiometry :
As 1 mole of [tex]O_2[/tex] is given by = 2 moles of [tex]KMnO_4[/tex]
Thus 0.05 moles of [tex]O_2[/tex] is given by =[tex]\frac{2}{1}\times 0.05=0.10moles[/tex] of [tex]KMnO_4[/tex]
Mass of [tex]KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g[/tex]
Thus 15.8 g of [tex]KMnO_4[/tex] will be required to produce 1.60 grams of [tex]O_2[/tex]
