For the reaction represented by the equation 2KClO3 → 2KCl + 3O2, how many grams of potassium chlorate are required to produce 160 g of oxygen? Type the numeric value only.

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