The grams of potassium chlorate that are required to produce 160 g of oxygen is 408.29 grams
<u><em>calculation</em></u>
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
The question is incomplete. Complete question is attached below
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Answer:
Equilibrium constant (K) is defined as the ratio of concentration of product to concentration of reactant.
For the reaction: 2H2O(g) ↔ 2H2(g) + O2(g)
Equlibrium constant =
![\frac{[H_2]^2[O_2]}{[H_20]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BH_20%5D%5E2%7D%20)
[ ] bracket in above expression indicates concentration.
Explanation:
vertical columns and horizontal rows, hope it helps