# Question #7b9b5

Mar 30, 2017

21.31 g of HF

#### Explanation:

To solve this question, one must use Moles to calculate the masses of the different compounds.

First we must determine the molecular weight of $H F$ and $S i {O}_{2}$. This can be done quickly by counting the masses of the elements. In your question, you have the elements (Si, F, H and O) defined in g/mol, which is not true. This unit should be $u$.
MW=molecular weight

MW $H F$ => $1.008 u \left(H\right) + 19.00 u \left(F\right)$=20.008 g/mol.
MW $S i {O}_{2}$ => $28.09 u \left(S i\right) + 2 \cdot 16.00 u \left(O\right)$ = 60.09 g/mol.
The mass of O should be multiplied by factor two since you have 2 oxygen atoms in $S i {O}_{2}$.

Then we proceed to calculate the number of moles of the reactant $S i {O}_{2}$. To calculate moles from grams, one must devide the amount of grams with the MW.
We have got 16.0 gram of $S i {O}_{2}$, therefore we have
$\frac{16.0}{60.09} = 0.266 \ldots .$ moles of $S i {O}_{2}$

We now look at the mole ratio between $S i {O}_{2}$ and $H F$, which is 1 : 4
To get the amount of moles that will react with 0.266... moles of $S i {O}_{2}$, we have to multiply it by 4 and devide it by 1. Therefore we get

$\frac{0.266 \ldots . \cdot 4}{1} = 1.07 \ldots$ moles of HF

The question asks us to calculate the mass of HF that will react with $S i {O}_{2}$, therefore we must calculate the mass of HF from the moles of HF. This can be done by doing the revers step we did before! We now muliply (instead of devide) the moles with the MW of HF.

$1.07 \ldots \cdot 20.008 = 21 , 3. . .$gram HF

Finally, we must give the answer in the appropriate amount of significant digits (the lowest one given has 4 significant digits):

21.31 g of HF

And that's it! Try to calculate it by yourself again so you get the drill!

For future calculations ALWAYS determine the moles! Most of the questions can be solved that way.