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# How many grams are there in $4.50{\text{ }}moles$ of $Ba{(N{O_2})_2}$ ?

Last updated date: 13th Jun 2024
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Hint: The simple formula is $grams{\text{ }} = {\text{ }}moles{\text{ }} \times {\text{ }}molar{\text{ }}mass$ you recognize the moles is $4.50$ and you can figure out the molar mass of $Ba{(N{O_2})_2}$ starting the periodic table.
You have to add the molar mass of every element. .

Complete step by step answer:
$number{\text{ }}of{\text{ }}moles{\text{ }}\left( n \right){\text{ }} = {\text{ }}4.50{\text{ }}moles$
$mass{\text{ }}\left( m \right){\text{ }} = \;n \times M\;$ [This is what you remain trying to discover]
$molar{\text{ }}mass{\text{ }}\left( M \right){\text{ }} = {\text{ }}?$
Your first step is to find on view what is the molar mass $\left( M \right)$ of $Ba{(N{O_2})_2}$ .
To discover a molar mass of a compound, opportunity the compound up into component parts so you could see what elements are elaborate.
If you break $Ba{(N{O_2})_2}$ up, you can obviously see that it is composed of $Ba,{\text{ }}N$ and $O$ .
You came to find the molar mass of each element elaborate. You can classify the molar mass straight away if you appear into the periodic table.
$Ba{\text{ }} = \;137.3gmo{l^{ - 1}}$
$N{\text{ }} = \;14.0 \times 2$
since there is a number $\;2$ after the closing bracket so $28.0gmo{l^{ - 1}}$
$O{\text{ }} =$ There are $2{\text{ }}O$ atoms classified the bracket and doubled because of the $\;2$ outside the closing bracket, so there are $4{\text{ }}O$ atoms overall.
$16.0gmo{l^{ - 1}} \times 4 = 64.0gmo{l^{ - 1}}$
So to become the molar mass of $Ba{(N{O_2})_2}$ , add the molar mass of the elements up. $137.3 + 28.0 + 64.0\;$ and that will provide you $229.3gmo{l^{ - 1}}$ . Then the molar mass of $Ba{(N{O_2})_2}$ is $229.3gmo{l^{ - 1}}$ .
$n{\text{ }} = {\text{ }}4.50{\text{ }}moles$
$m{\text{ }} = {\text{ }}?$
$M{\text{ }} = {\text{ }}229.3\;gmo{l^{ - 1}}$
Step 2, you can at present find the mass of $Ba{(N{O_2})_2}$ . To find the mass, you necessity to multiply the number of moles $\left( n \right)$ with molar mass $\left( M \right),$ in other words, $m = n \times M$
$M{\text{ }} = 4.50 \times 229.3 = 1031.85$

Then, there are $1031.85{\text{ }}grams$ in $4.50{\text{ }}moles$ of $Ba{(N{O_2})_2}$

Note: In chemistry, the formula weight is a capacity computed by multiplying the atomic weight (in atomic mass components) of every element in a chemical formula by the sum of atoms of that element extant in the formula, formally adding all of these products together.
Definition molar mass starts with units of grams per mole $\left( {g/mol} \right).$ Once computing molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is basically the weight in atomic mass units of all the atoms in an assumed formula.
By means of the chemical formula of the compound and the periodic table of elements, we can increase up the atomic weights and compute the molecular weight of the substance.