SHOW WORK PLEASE When heated, KClO3 decomposes into KCl and O2. 2KClO3 2KCl + 3O

Question

SHOW WORK PLEASE

When heated, KClO3 decomposes into KCl and O2. 2KClO3 2KCl + 3O2 If this reaction produced 88.9 g of KCl, how much O2 was produced (in grams)? Number 38.19 g O2 SHOW WORK PLEASE The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 3.876 g sample of ether was combusted in an oxygen rich environment to produce 9.205 g of CO2(g) and 4.710 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of ether. CHO

Explanation / Answer

1)

In the combustion all the H in ether will be concerted to H2O and C will be converted to CO2.

Amount of H2O produced = 4.71g

Amount of CO2 produced = 9.205g

Molar mass

H2O = 18 g/mol

CO2 = 44 g/mol

Amount of H2O moles produced = 4.71/18 = 0.2617 mol

Amount of CO2 produced = 9.205/44 = 0.2092 mol

Let us say the emperical formula is CxHyOz

We need 2mol of H in ether to produce 1 mol of H2O.

We can say we had;

C = 0.2092 = x

H = 0.2617*2 = 0.5234 = y

in the ether.

The mass of ether is 3.876g.

We can say;

12x+1y+16z = 3.876

12*0.2092+1*0.5234+16z = 3.876

z = 0.0526

x:y:z = 0.2092:0.5234:0.0526 = 0.2092/0.0526:0.5234/0.0526:0.0526/0.0526 = 3.977:9.95:1 = 4:10:1

So the emperical formular is C4H10O

2)

Mole mass

KCl = 74.55g/mol

O2 = 32g/mol

Amount of KCl produced = 88.9g = 88.9/74.55 mol = 1.192 mol

Mole ratio

KCl:O2 = 2:3

Amount of O2 produced = 1.192/2*3 = 1.788g

Amount of O2 produced = 1.788*32 = 57.216g

So the answer is 57.216g

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