Question

The ratio of oxygen atoms in 12.25 g of potassium chlorate and 1.06 g of sodium carbonate is _______.

Answer 1

potassium chlorate atomic weight = 122.55 gm

number of mole potassium chlorate in 12.25gm = $\frac{12.25}{122.55}$= 0.01 mole

number of oxygen atoms in that amount of $KCl{O}_3$ is = $3\times 0.01moles\times 6.02\times {10}^{23}$atoms / mole = $1.806\times {10}^{22}$ atoms

sodium carbonate atomic weight = 106 gm

number of mole sodium carbonate in 1.06 gm = $1.06/106=0.01mole$

number of oxygen atoms in that amount of $Na2C{O}_3$is = $3\times 0.01moles\times (6.02\times {10}^{23}$ atoms / mole) = $1.806\times {10}^{22}$ atoms

answer = $\frac{\text{number of oxygen atoms in}KCl{O}_3}{\text{number of oxygen atoms in}N{a}_{2}C{O}_3}$

answer = 1