In a book, which contains only stoichiometry problems, I came across this formula.
For any given compound $\ce{M_xN_y}$, $x \text{ Moles of N} = y \text{ moles of M}$
According to this rule if I take $\ce{H2O}$ then, $x = 2 \text{ and } y = 1$.
So by the above formula,
$2 \text{ moles of Oxygen} = 1 \text{ mole of Hydrogen}$
But surely,
$2 * \mathrm{N_A} \not= 1 * \mathrm{N_A}$, where $\mathrm{N_A}$ is Avogadro's number.
However, in an example, the author has used this formula and successfully solved a problem? So, if it's correct, can anyone prove this and point out where my contradiction is wrong.
Example as requested:
What weight of oxygen will react with $1$g of calcium?($\ce{Ca}= 40\mathrm{g \over mol}$)
Answer by author : Since all the atoms of $\ce{Ca}$ have changed into $\ce{CaO}$, the amount of $\ce{Ca}$ in $\ce{CaO}$ is $1$g. Now from the formula of $\ce{CaO}$, we have,
$\text{Moles of Ca = Moles of O}$
$$\frac{\text{mass of Ca}}{\text{atomic mass of Ca}}= \frac{\text{mass of O}}{\text{atomic mass of O}}$$
therefore, $\text{mass of O} = {1\over40} * 16 = 0.4$g