Question

Maximum number of moles of potash alum ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$can be made from sample of aluminium-containing $5.4$g of aluminium (considering $100$% yield) is?
A. ${10^{ - 2}}\,{\text{mole}}$
B. $0.5\, \times {10^{ - 2}}\,{\text{moles}}$
C. $0.2\,{\text{moles}}$
D. $0.1\,\,{\text{mole}}$

Answer 1

first we will determine how many mole aluminium is forming one mole alum. We will determine the mole of aluminium from the given amount. After this, by comparing the number of moles of aluminium and alum with the amount of aluminium present, we can determine the mole of alum.

Complete Step by step answer: The formula of potash alum is ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$. By the formula it is clear that two mole aluminium is required for the preparation of one mole of alum.
We will determine the mole of aluminium as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of sodium oxide aluminium is $27{\text{g/mol}}$
Substitute $27{\text{g/mol}}$for molar mass and $5.4$g for mass of aluminium.
${\text{Mole}}\,{\text{ = }}\,\dfrac{{5.4\,{\text{g}}}}{{27\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mole}}\,{\text{ = }}\,0.2$
According to the formula of alum, two-mole aluminium is giving one mole of alum so, the mole of alum, given by the $0.2$ mole of aluminium is,
$2\,{\text{mol}}\,\,{\text{Al}}\, = \,1\,\,{\text{mol}}\,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$
$\Rightarrow 0.2\,{\text{mol}}\,\,{\text{Al}}\, = \,0.1\,\,{\text{mol}}\,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$
So, $0.1\,$ mole of potash alum ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$can be made from sample of aluminium-containing $5.4$g of aluminium.

Therefore, option (D) $0.1\,\,{\text{mole}}$ is correct.

Note: In place of moles, we can compare the gram weight also. The gram weight of aluminium is $27$so, weight of two mole will be$54$ and the gram weight of alum is $474$. According to formula of alum, $54$ gram of aluminium is giving$474$gram of alum so, $5.4$gram will give,
$54\,{\text{g}}\,\,{\text{Al}}\, = \,474\,{\text{g}}\,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$
$\Rightarrow 5.4\,{\text{g}}\,\,{\text{Al}}\, = \,47.4\,{\text{g}}\,\,{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{A}}{{\text{l}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}$
So, the mole of alum will be,
${\text{Mole}}\,{\text{ = }}\,\dfrac{{47.4\,{\text{g}}}}{{474\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mole}}\,{\text{ = }}\,0.1$
Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary.