
How many grams of ammonia is obtained when 32.6g of ammonium chloride reacts with calcium hydroxide during the laboratory preparation of ammonia?
$2N{H_4}Cl + Ca{\left( {OH} \right)_2} \to CaC{l_2} + 2{H_2}O + 2N{H_3}$
$N = 14,\;H = 1,\;O = 16,\;S = 32,\;Cl = 35.5$
A.10.36 gm
B.53.9 gm
C.42.95 gm
D.5.18 gm
Answer
561k+ views
Hint: We can use the stoichiometry of balanced reaction to have many information about that reaction. We could find their number of moles relation, mass relation and also the volume relation.
Complete step by step solution:
As we said we can get a lot of information from the above balanced reaction. Here we can ammonium chloride reacting with calcium hydroxide to form ammonium and calcium chloride along with water.
Now, let us look at what all the information we can get from the reaction. It says when 2 moles of ammonium chloride reacted with 1 mole of calcium hydroxide to form 2 moles of ammonium, 1 mole of calcium chloride and 2moles of water.
Now, we can check the molar mass of ammonium chloride
$molar\;mass\;N{H_4}Cl = \;14 \times 1 + 1 \times 4 + 35.5 \times 1 = 53.5g$
This means that 1 mole of $N{H_4}Cl$ contains 53.5g of $N{H_4}Cl$
So we understood that, we need 2 moles of $N{H_4}Cl$ to produce 2 moles ammonia, hence we can say that 1 mole of $N{H_4}Cl$ is required to form ammonia when reacted with calcium hydroxide.
$Molar\;mass\;of\;N{H_3} = 14 \times 1 + 1 \times 3 = 17g$
Also we now say that 53.5g of $N{H_4}Cl$ produces 17g of $N{H_3}$
Hence, 1g $N{H_4}Cl$ produces $\dfrac{{17}}{{53.5}}g$ of $N{H_3}$
In question it is given that 32.6g of ammonium chloride is reacted with calcium hydroxide.
Therefore, 32.6g of $N{H_4}Cl$ produces $32.6 \times \dfrac{{17}}{{53.5}}g$ of $N{H_3}$
i.e., we can say that 32.6g of $N{H_4}Cl$ produces 10.6g of $N{H_3}$
Hence, option (a) is correct
Note: If the reaction is not balanced then we cannot apply stoichiometry. Therefore, whenever we are applying the stoichiometry in a reaction we have to first make sure that it is balanced and if not balance the reaction first then apply the stoichiometry principles to find the required information.
Complete step by step solution:
As we said we can get a lot of information from the above balanced reaction. Here we can ammonium chloride reacting with calcium hydroxide to form ammonium and calcium chloride along with water.
Now, let us look at what all the information we can get from the reaction. It says when 2 moles of ammonium chloride reacted with 1 mole of calcium hydroxide to form 2 moles of ammonium, 1 mole of calcium chloride and 2moles of water.
Now, we can check the molar mass of ammonium chloride
$molar\;mass\;N{H_4}Cl = \;14 \times 1 + 1 \times 4 + 35.5 \times 1 = 53.5g$
This means that 1 mole of $N{H_4}Cl$ contains 53.5g of $N{H_4}Cl$
So we understood that, we need 2 moles of $N{H_4}Cl$ to produce 2 moles ammonia, hence we can say that 1 mole of $N{H_4}Cl$ is required to form ammonia when reacted with calcium hydroxide.
$Molar\;mass\;of\;N{H_3} = 14 \times 1 + 1 \times 3 = 17g$
Also we now say that 53.5g of $N{H_4}Cl$ produces 17g of $N{H_3}$
Hence, 1g $N{H_4}Cl$ produces $\dfrac{{17}}{{53.5}}g$ of $N{H_3}$
In question it is given that 32.6g of ammonium chloride is reacted with calcium hydroxide.
Therefore, 32.6g of $N{H_4}Cl$ produces $32.6 \times \dfrac{{17}}{{53.5}}g$ of $N{H_3}$
i.e., we can say that 32.6g of $N{H_4}Cl$ produces 10.6g of $N{H_3}$
Hence, option (a) is correct
Note: If the reaction is not balanced then we cannot apply stoichiometry. Therefore, whenever we are applying the stoichiometry in a reaction we have to first make sure that it is balanced and if not balance the reaction first then apply the stoichiometry principles to find the required information.
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