2KCIO3 --- 2 KCI + 3O2
I want to produce 5.7 mol of O2 . How many grams of KCIO3 should i start with?
3.8g KCIO3
698.54g KCIO3
5.7g KCIO3
465.69g KCIO3

465.69g KCIO3

Explanation:

See the stoichiometry in the reaction:

We can propose

3 moles of oxgen are made of 2 moles of chlorate

Therefore 5.70 moles of O₂ will be made by (5.70. 2) / 3 = 3.8 moles of chlorate.

We convert the moles to mass: 3.8 mol . 122.55 g/ 1 mol = 465.69g KCIO3

is the mass to use in the begining

We have to start with 465.69 grams of KClO3 (option 4 is correct)

Explanation:

Step 1: Data given

Moles of O2 = 5.7 moles

Molar mass KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KCIO3 → 2KCI + 3O2

Step 3: Calculate moles KClO3

For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

For 5.7 moles O2 we nee 2/3 * 5.7 = 3.8 moles KClO3

Step 4: Calculate mass of KClO3

Mass KClO3 = moles KClO3 * molar mass KClO3

Mass KClO3 = 3.8 moles * 122.55 g/mol

Mass KClO3 = 465.69 grams

We have to start with 465.69 grams of KClO3 (option 4 is correct)

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2