# Question 578d1

Feb 1, 2016

$\text{moles(H"_2")} = 1.28$ $\text{moles}$

#### Explanation:

First we acknowledge that

${\text{moles" = "mass /g"/"M}}_{r}$

Consulting the data with which we have been provided, the most obvious progression would be to calculate the moles of $\text{H"_2"O}$ produced (we have been given the mass, and we can easily calculate the ${\text{M}}_{r}$ of water).

$\text{M"_r"(H"_2"O)} = 2 \times 1 + 16 = 18$

N.B. you would be wise to keep this number in your head in future, given the prominence of water in this discipline.

$\therefore$ $\text{moles(H"_2"O)} = \frac{23}{18} = 1.2 \dot{7}$ $\text{moles}$

In this question we must simply find the number of moles of hydrogen gas we required to make this much water, so we need only consult the reaction's mole ratio. Let's look at the formula, which I have decided to construct for the preparation of one mole of water from hydrogen and oxygen gases:

$\text{H"_2 + 1/2 "O"_2 -> "H"_2"O}$

Notice that we could have doubled up these coefficients for a slightly friendlier-looking equation, but expressing the reaction in this way makes our mole ratio much more obvious and, if you're given marks in an exam for showing working, you don't need to employ a useless $\frac{2}{2}$ in your calculations.

To conclude:

$\text{moles(H"_2")" = "moles(H"_2"O)} = 1.2 \dot{7} = 1.28$ $\text{moles to 3 significant figures}$

Feb 1, 2016

$\text{1.3 mol H"_2}$ were needed to produce $\text{23 g H"_2"O}$, if $\text{O"_2}$ is present in excess.

#### Explanation:

${\text{2H"_2 + "O}}_{2}$$\rightarrow$$\text{2H"_2"O}$

Determine molar mass of $\text{H"_2"O}$.

The molar mass of a molecule is determined by multiplying the subscripts times the atomic masses of the elements on the periodic table in g/mol.

$\text{H"_2"O} :$$2 \times 1.00794 \text{g/mol" + 15.9994"g/mol"=18.01528"g/mol}$

Determine the moles of water that was produced by dividing the mass of water produced by its molar mass.

23cancel("g H"_2"O")xx(1"mol H"_2"O")/(18.01528cancel("g H"_2"O"))="1.2767 mol H"_2"O"

I am leaving a few extra digits to reduce rounding errors. The final answer will be rounded to two significant figures.

Next determine the moles of $\text{H"_2}$ needed by multiplying the moles of $\text{H"_2"O}$ times the mole ratio $\left(2 \text{mol H"_2)/(2"mol H"_2"O}\right)$ from the balanced equation.

1.2767cancel("mol H"_2"O")xx(cancel(2)"mol H"_2)/(cancel(2)cancel("mol H"_2"O"))="1.3 mol H"_2" rounded to two significant figures

This problem can be worked out all at once by combining the steps.

23cancel("g H"_2"O")xx(1cancel("mol H"_2"O"))/(18.01528cancel("g H"_2"O"))xx(cancel2"mol H"_2)/(cancel2cancel("mol H"_2"O"))="1.3 mol H"_2"# rounded to two significant figures