__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
the element chlorine is represented by the symbol Cl
The answer should be E,C,F,B,A,D in that order.
I will present a simple reaction so we can do this conversion:
2H₂ + O₂ → 2H₂O
We will assume we have 32 g of O₂ and we want to find the amount of water, assuming this reaction goes to completion. We must first convert the initial mass to moles, which we do using the molar mass in units of g/mol. The molar mass of O₂ is 32 g/mol.
32 g O₂ ÷ 32 g/mol = 1 mole O₂.
Now that we have moles of oxygen, we use the molar coefficients to find the ratio of water molecules to oxygen molecules. We can see there are 2 moles of water for every 1 mole of oxygen.
1 moles O₂ x (2 mol H₂O/ 1 mol O₂) = 2 moles H₂O
Now that we have the moles of water, we can convert this amount into grams using the molar mass of water, which is 18 g/mol.
2 moles H₂O x 18 g/mol = 36 g H₂O
Now we have successfully converted the mass of one molecule to the mass of another.
