Consider the following reaction:
KClO3(s) KCl(s)+ O2(g) (unbalanced)

8.06 moles of potassium chlorate will yield how many:
moles of oxygen and grams of KCl

1. 12.09 mole

2. 600.47g

Explanation:

First, we'll begin by writing a balanced equation for the reaction. The balance equation is given below:

2KClO3(s) —> 2KCl(s)+ 3O2(g)

1. From the balanced equation above,

2 moles of potassium chlorate (KClO3) produced 3 moles of oxygen (O2) .

Therefore, 8.06 moles of potassium chlorate (KClO3) will produce = (8.06x3)/2 = 12.09 moles of oxygen (O2).

From the calculations above, 8.06 moles of potassium chlorate (KClO3) will produce 12.09 moles of oxygen (O2)

2. From the balanced equation above,

2 moles of potassium chlorate (KClO3) produced 2 moles of potassium chloride (KCl).

Therefore, 8.06 moles of potassium chlorate (KClO3) will also produce 8.06 moles of potassium chloride (KCl).

Now, we need to convert 8.06 moles of potassium chloride (KCl) to grams in order to obtain the desired result. This is illustrated below:

Molar Mass of KCl = 39 + 35.5 = 74.5g/mol

Mole of KCl = 8.06 moles

Mass of KCl =?

Mass = number of mole x molar Mass

Mass of KCl = 8.06 x 74.5

Mass of KCl = 600.47g

Therefore, 8.06 moles of potassium chloride (KCl) will produce 600.47g of potassium chloride (KCl)

the answer is na (sodium)

it would be a gas formation