# If 3.00 moles of H_2O are produced, how many grams of hydrogen gas are used?

Apr 26, 2016

If the equation is
${H}^{+}$ + ${O}^{-} 2$ $\implies$ ${H}_{2} O$

and if you balance it,
$4 {H}^{+}$ + $2 {O}^{-} 2$ $\implies$ $2 {H}_{2} O$

Assume that:
n = number of moles
m = mass of substance
M = molar mass (equivalent to atomic weight on the periodic table)

$n = m \div M$

The mole ratio between $H : {H}_{2} O$ is $4 : 2$. If you simplify the mole ratio into simplest form, it becomes $2 : 1$

So, if 1 mole of ${H}_{2} O$ gives you 2 moles of ${H}^{+}$,

then 3.00 moles of ${H}_{2} O$ must give you:
$\left[3.00 m o l \times 2\right]$ = 6.00 moles of ${H}^{+}$

You now know that Hydrogen gas has 6.00 moles. (n)

Your next step is to find the molar mass (M) of hydrogen (${H}^{+}$). If you look into your periodic table, the molar mass of (${H}^{+}$) is 1.00 g/mol.

Since you have found your number of moles (n) and molar mass (M) for hydrogen, your final step is to determine the mass (m).

Looking back at the $n = m \div M$ formula, to find mass of substance (m), you need to flip it and it becomes $m = M \times n$

The mass of hydrogen gas is:
m = 1.00 g/mol $\times$ 6.00 moles = 6.00 grams of ${H}^{+}$ gas.

Therefore 6.00 grams of hydrogen gas is being used.