How many moles of sodium hydroxide will be produced from the complete reaction of \[15.0{\text{ }}g\] sodium in the reaction \[2Na{\text{ }} + {\text{ }}2{H_2}O\;\, \to \,\;2NaOH{\text{ }} + {\text{ }}{H_2}\] ?
Answer
588.9k+ views
Hint: A mole is a unit measurement for an amount of substance in the international system of units i.e., SI unit. A mole of a particle or a mole of a substance is defined as \[6.02214076 \times {10^{23}}\] of a chemical unit, that can be ions, atoms, molecules, etc. Originally it was defined as, the number of atoms in \[12{\text{ }}g\] of carbon-12.
Complete step-by-step answer:
Now, the definition is just one mole is \[6.02214076 \times {10^{23}}\] of a chemical unit.
In a chemical equation, on the left side it shows reactant and on the right side the product. But when it comes to the balanced chemical equation, it is shown in the terms of moles.
Such that, how many moles are required in this reaction. Stoichiometry is the sturdy relationship of quantities of substances in a chemical reaction.
To calculate the number of moles, we need use the equation;
Number of moles, \[n\, = \,\dfrac{{mass}}{{molecular\,mass}}\]
The given mass of sodium= \[15.0{\text{ }}g\]
The molecular mass of sodium = \[23{\text{ }}g/mol\]
Substitute the values in above equation,
\[n\, = \,\dfrac{{15.0\,\,g}}{{23\,\,g/mol}}\]
We get,
\[n\, = \,0.652\,\,moles\]
Now, let’s see the chemical equation;
\[2Na{\text{ }} + {\text{ }}2{H_2}O\;\, \to \,\;2NaOH{\text{ }} + {\text{ }}{H_2}\]
It says that \[1\] mole of sodium produces \[1\] mole of sodium hydroxide.
It means when \[1\] mole of sodium is reacted it forms \[1\] mole of sodium hydroxide.
Therefore, by seeing the stoichiometry of this reaction,
We observe,
\[2\] mole of sodium is reacted; it forms \[2\] mole of sodium hydroxide.
So, \[n\, = \,0.652\,\,moles\] of sodium will produce= \[ = \,\,0.652\, \times \,\,\dfrac{2}{2}\] \[ = \,\,0.652\,\]moles of \[NaOH\]
Hence, the number of moles of \[NaOH\] produced if \[0.652\,\] moles.
Note: If a compound is having the mass, we can calculate the moles by using the formula of moles, either it is for one molecule or the compound. If a compound is having \[A\] and \[B\] molecules then the sum of the moles will be \[1\].
\[{n_A}\, = \,\dfrac{{mass\,of\,A}}{{molecular\,mass\,of\,A}}\]
\[{n_B}\, = \,\dfrac{{mass\,of\,B}}{{molecular\,mass\,of\,B}}\]
\[{n_A}\, + {n_B}\, = \,1\]
Complete step-by-step answer:
Now, the definition is just one mole is \[6.02214076 \times {10^{23}}\] of a chemical unit.
In a chemical equation, on the left side it shows reactant and on the right side the product. But when it comes to the balanced chemical equation, it is shown in the terms of moles.
Such that, how many moles are required in this reaction. Stoichiometry is the sturdy relationship of quantities of substances in a chemical reaction.
To calculate the number of moles, we need use the equation;
Number of moles, \[n\, = \,\dfrac{{mass}}{{molecular\,mass}}\]
The given mass of sodium= \[15.0{\text{ }}g\]
The molecular mass of sodium = \[23{\text{ }}g/mol\]
Substitute the values in above equation,
\[n\, = \,\dfrac{{15.0\,\,g}}{{23\,\,g/mol}}\]
We get,
\[n\, = \,0.652\,\,moles\]
Now, let’s see the chemical equation;
\[2Na{\text{ }} + {\text{ }}2{H_2}O\;\, \to \,\;2NaOH{\text{ }} + {\text{ }}{H_2}\]
It says that \[1\] mole of sodium produces \[1\] mole of sodium hydroxide.
It means when \[1\] mole of sodium is reacted it forms \[1\] mole of sodium hydroxide.
Therefore, by seeing the stoichiometry of this reaction,
We observe,
\[2\] mole of sodium is reacted; it forms \[2\] mole of sodium hydroxide.
So, \[n\, = \,0.652\,\,moles\] of sodium will produce= \[ = \,\,0.652\, \times \,\,\dfrac{2}{2}\] \[ = \,\,0.652\,\]moles of \[NaOH\]
Hence, the number of moles of \[NaOH\] produced if \[0.652\,\] moles.
Note: If a compound is having the mass, we can calculate the moles by using the formula of moles, either it is for one molecule or the compound. If a compound is having \[A\] and \[B\] molecules then the sum of the moles will be \[1\].
\[{n_A}\, = \,\dfrac{{mass\,of\,A}}{{molecular\,mass\,of\,A}}\]
\[{n_B}\, = \,\dfrac{{mass\,of\,B}}{{molecular\,mass\,of\,B}}\]
\[{n_A}\, + {n_B}\, = \,1\]
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