Question

Given the following reaction ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {\text{AgBr}} \to {\text{N}}{{\text{a}}_{\text{3}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_{\text{2}}}} \right] + {\text{NaBr}}$. How many moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ are needed to react completely with $42.7{\text{ g}}$ of ${\text{AgBr}}$?

Answer 1

Hint: To solve this first balance the reaction. Then from the reaction stoichiometry calculate the calculate the mole ratio between ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and ${\text{AgBr}}$. From the mole ratio, calculate the number of moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ that are needed to react completely with $42.7{\text{ g}}$ of ${\text{AgBr}}$.

Complete solution:
We are given the reaction as follows:
${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {\text{AgBr}} \to {\text{N}}{{\text{a}}_{\text{3}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_{\text{2}}}} \right] + {\text{NaBr}}$
The given reaction is not balanced. The balanced reaction is as follows:
${\text{2N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {\text{AgBr}} \to {\text{N}}{{\text{a}}_{\text{3}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_{\text{2}}}} \right] + {\text{NaBr}}$
Calculate the number of moles of ${\text{AgBr}}$ in $42.7{\text{ g}}$ of ${\text{AgBr}}$ as follows:
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $42.7{\text{ g}}$ for the mass of ${\text{AgBr}}$ and $187.77{\text{ g/mol}}$ for the molar mass of ${\text{AgBr}}$. Thus,
${\text{Number of moles of AgBr}} = \dfrac{{42.7{\text{ g}}}}{{187.77{\text{ g/mol}}}}$
${\text{Number of moles of AgBr}} = 0.2274{\text{ mol}}$
Thus, the number of moles of ${\text{AgBr}}$ in $42.7{\text{ g}}$ of ${\text{AgBr}}$ are $0.2274{\text{ mol}}$.
Calculate the number of moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ needed to react with $0.2274{\text{ mol}}$ of ${\text{AgBr}}$ as follows:
From the reaction stoichiometry, we can say that
${\text{2 mol N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} \equiv 1{\text{ mol AgBr}}$
Thus,
${\text{Number of moles of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 0.2274{\text{ mol AgBr}} \times \dfrac{{2{\text{ mol N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}{{1{\text{ mol AgBr}}}}$
${\text{Number of moles of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = 0.445{\text{ mol N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
Thus, the number of moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ needed to react with $0.2274{\text{ mol}}$ of ${\text{AgBr}}$ are $0.445{\text{ mol}}$.
Thus, the moles of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ needed to react completely with $42.7{\text{ g}}$ of ${\text{AgBr}}$ are $0.445{\text{ mol}}$.

Note:Remember that the balanced chemical equation for the given reaction must be written correctly. Incorrect or unbalanced chemical equations can lead to incorrect number of moles which can lead to incorrect mass of the element.