Question

If $1\dfrac{1}{2}$ moles of oxygen combine with $Al$ to form $A{l_2}{O_3}$, the weight of $Al$ used in the reaction is $\left( {Al = 27} \right)$.
(a)$27g$
(b)$54g$
(c)$40.5g$
(d)$81g$

Answer 1

Hint: We recognize that the number of moles in a given amount of any substance is adequate to the grams of the substance divided by its molecular weight.
We can write the mathematical expression for mole as,
${\text{Mole}} = \dfrac{{{\text{weight of substance}}}}{{{\text{Molecular weight}}}}$

Complete step by step solution:
Aluminium oxide is formed when aluminium is reacted with oxygen, the chemical reaction is,
$4Al\left( s \right) + 3{O_2}\left( g \right) \to 2A{l_2}{O_3}\left( s \right)$
Three moles of oxygen gas reacts with four moles of aluminium.
So, half mole of oxygen gas will react with aluminium metal.
The mass of aluminium can be calculated as,
${\text{Mole}} = \dfrac{{{\text{weight of substance}}}}{{{\text{Molecular weight}}}}$
We know that the molar weight of aluminium \[ = 27g/mol\]
Half moles of oxygen reacts with $ = \dfrac{4}{3} \times \dfrac{3}{2} = 2mole$
Substituting the known values in the above equation,
$2 = \dfrac{{{\text{Weight of the substance}}}}{{27g}}$
Weight of the substance$ = 27g \times 2$
Weight of the substance$ = 54g$
Therefore, the mass of aluminium worn in the reaction is \[54grams\].
Therefore, option (b) is correct.

Note:
We apprehend that a mole magnitude relation could be a ratio between the numbers of moles of any 2 species concerned in an exceedingly chemical reaction.
The given equation is,
${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$
Six gram of hydrogen reacts with twenty eight grams of nitrogen to offer ammonia.
The total mass of ammonia $ = 2 \times 14 + 2\left( 3 \right) = 34$
One gram of nitrogen reacts with one gram of hydrogen to give $\dfrac{{34}}{{28}} \times 1g$ ammonia.
When $2 \times {10^3}\,g$ of Dinitrogen reacts with $1 \times {10^3}g$ of hydrogen to give,
$\dfrac{{34}}{{28}} \times 2 \times {10^3}g = 2428.57g$
The mass of ammonia produced if $2 \times {10^3}g$ Dinitrogen reacts with $1 \times {10^3}g$ of hydrogen is $2428.57g$.
Hence nitrogen is the limiting agent.
Dinitrogen is that the limiting chemical agent and gas is the excess reagent. Hence, dihydrogen can stay unreacted.
Amount of hydrogen that remains unreacted.
Six gram of hydrogen reacts with twenty eight grams of nitrogen to offer ammonia. Hence $2 \times {10^3}g$ require $ = \dfrac{6}{{28}} \times 2 \times {10^3} = 428.5g$
Amount of hydrogen that remains unreacted$ = 1 \times {10^3}g - 428.5g = 571.5g$.