A) (n/2)g(B) 100g(C) (100/n)g(D) 100ngC-6s How many moles of potassium

1.	A) (n/2)g(B) 100g(C) (100/n)g(D) 100ngC-6s How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at N(A) mol2(B) 1mol(D)ëĽź(C)mol34
Answer» The production of oxygen from per chlorate is as follows: 2KClO3(s) -> 2KCl (s) + 3O2(g) so, here 2 moles of per chlorate is used to produce 3 moles of oxygen gas. So, 1 mole of oxygen gas will need = 2/3 moles of per chlorate. Now we know that 1 mole of oxygen gas at STP = 22.4 L So, 112 L oxygen gas = 112/22.4 =5 Moles of oxygen gas. So, 5 moles (112L) of oxygen gas will be produced by = 5 x 2/3 moles of per chlorate = 3.33 moles

A) (n/2)g(B) 100g(C) (100/n)g(D) 100ngC-6s How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at N(A) mol2(B) 1mol(D)ëĽź(C)mol34

Answer»

The production of oxygen from per chlorate is as follows:

2KClO3(s) -> 2KCl (s) + 3O2(g)

so, here 2 moles of per chlorate is used to produce 3 moles of oxygen gas.

So, 1 mole of oxygen gas will need = 2/3 moles of per chlorate.

Now we know that 1 mole of oxygen gas at STP = 22.4 L

So, 112 L oxygen gas = 112/22.4 =5 Moles of oxygen gas.

So, 5 moles (112L) of oxygen gas will be produced by = 5 x 2/3 moles of per chlorate = 3.33 moles