Question

2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2

Answer 1

Answer: 15.8 g of $KMnO_4$ will be required to produce 1.60 grams of $O_2$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles$

$2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$

According to stoichiometry :

As 1 mole of $O_2$ is given by = 2 moles of $KMnO_4$

Thus 0.05 moles of $O_2$ is given by =$\frac{2}{1}\times 0.05=0.10moles$  of $KMnO_4$

Mass of $KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g$

Thus 15.8 g of $KMnO_4$ will be required to produce 1.60 grams of $O_2$