The grams of potassium chlorate that are required to produce 160 g of oxygen is 408.29 grams
calculation
2 KClO₃→ 2 KCl + 3O₂
Step 1: find the moles of O₂
moles = mass÷ molar mass
from periodic table the molar mass of O₂ = 16 x2 = 32 g/mol
moles = 160 g÷ 32 g/mol = 5 moles
Step2 : use the mole ratio to determine the moles of KClO₃
from equation given KClO₃ : O₂ is 2:3
therefore the v moles of KClO₃ = 5 moles x 2/3 = 3.333 moles
Step 3: find the mass of KClO₃
mass= moles x molar mass
from periodic table the molar mass of KClO₃
= 39 + 35.5 + (16 x3) =122.5 g/mol
mass = 3.333 moles x 122.5 g/mol =408.29 grams
