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Calculate the number of atoms there are in a sample of aluminium that weighs $80,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .
Number of atoms in 3 mol aluminium $=3\times 6,022\times 10{}^{23}$
There are $18,069\times 10{}^{23}$ aluminium atoms in a sample of $80,94\phantom{\rule{2pt}{0ex}}\mathrm{g}$ .
So far, we have only discussed moles, mass and molar mass in relation to elements . But what happens if we are dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer is 'yes'. However, you need to remember that all your calculations will apply to the whole molecule . So, when you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound. Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid ( $\mathrm{HNO}{}_{3}$ ), it means you have $6,022\times {10}^{23}$ molecules of nitric acid in the sample. This also means that there are $6,022\times {10}^{23}$ atoms of hydrogen, $6,022\times {10}^{23}$ atoms of nitrogen and ( $3\times 6,022\times {10}^{23}$ ) atoms of oxygen in the sample.
In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.
e.g. ${\mathrm{N}}_{2}+3{\mathrm{H}}_{2}\to 2\mathrm{N}{\mathrm{H}}_{3}$
In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Calculate the molar mass of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ .
Hydrogen $=\mathrm{1,008}\phantom{\rule{2pt}{0ex}}\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ ; Sulphur = $=\mathrm{32,07}\phantom{\rule{2pt}{0ex}}\mathrm{g}\xb7\mathrm{mol}{}^{-1}$ ; Oxygen $=16\phantom{\rule{2pt}{0ex}}\mathrm{g}\xb7\mathrm{mol}{}^{-1}$
Calculate the number of moles there are in $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ of $\mathrm{MgCl}{}_{2}$ .
There are $10,5\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ of magnesium chloride in a $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ sample.
Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.
${\mathrm{BaCl}}_{2}+{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{BaSO}}_{4}+2\mathrm{HCl}$
If you have $2\phantom{\rule{2pt}{0ex}}\mathrm{g}$ of $\mathrm{BaCl}{}_{2}$ ...
According to the balanced equation, 1 mole of $\mathrm{BaCl}{}_{2}$ will react with 1 mole of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ . Therefore, if $0,0096\phantom{\rule{2pt}{0ex}}\mathrm{mol}$ of $\mathrm{BaCl}{}_{2}$ react, then there must be the same number of moles of $\mathrm{H}{}_{2}\mathrm{SO}{}_{4}$ that react because their mole ratio is 1:1.
(answer to 1)
According to the balanced equation, 2 moles of $\mathrm{HCl}$ are produced for every 1 mole of the two reactants. Therefore the number of moles of $\mathrm{HCl}$ produced is ( $2\times 0,0096$ ), which equals $0,0096\phantom{\rule{2pt}{0ex}}\mathrm{moles}$ .
(answer to 2)
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