Answer:
10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated
Explanation:
This is the balanced equation:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Ratio beteween the salt and oxygen is 2:3
Molar mass of KClO₃ = 122.55 g/m
Let's find out the moles of salt
Mass / Molar mass
26.4 g /122.55 g/m = 0.215 moles
So, this is the final rule of three:
If 2 moles of KClO₃ make 3 moles of oxygen
0.215 moles of KClO₃ make (0.215 .3) /2 = 0.323 moles of O₂ are produced
Molar mass O₂ = 32 g/m
Moles . molar mass = mass
0.323 m . 32g/m = 10.3 g
Omg i lost everything ugh
To do it again
1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol
2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol
3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol
4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol
5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g
6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g
7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g
8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g
I cant believe i had to do this all over
Answer: 323.61 g of 
will be produced
Explanation:
The given balanced chemical reaction is :
According to stoichiometry :
2 moles of 
require 1 mole of 
Thus 3.00 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product.
As 2 moles of give = 2 moles of
Thus 3.00 moles of give =
of
Mass of 
Thus 323.61 g of will be produced from the given moles of both reactants.
