Question

$Ca\left( s \right)+C{{l}_{2}}(g)\to CaC{{l}_{2}}\left( s \right)$
When 80g Ca (Molar mass is 40) is reacted with 213 g of $C{{l}_{2}}$ (molar mass is 71), one will have:
(A) 40 g Ca excess
(B) 71 g $C{{l}_{2}}$ excess
(C) 293 g $CaC{{l}_{2}}$ formed
(D) 133 g $CaC{{l}_{2}}$ formed
(E) 113 g $CaC{{l}_{2}}$ formed

Answer 1

Hint: To answer this question, firstly calculate the number of moles of the reactant species. Once you get that you can find the reactant present in excess. Convert the excess reactant present from moles to gram by multiplying it by its molar mass and this will give you the required answer.

Complete step by step solution:
To solve this, firstly let us convert the mass of the species to their number of moles. We can do this conversion by dividing the mass of the elements (that is given to us in the question) by molecular weights.
Firstly for calcium, it is given that 80g of calcium was used and we know that molar mass of calcium is 40 g/mol.
Therefore, we can write that-
Number of moles of calcium = $\dfrac{80g}{40g/mol}=2\text{ }mol$
Similarly, for chlorine, it is given that 213g of chlorine was reacted and the molar mass of chlorine is 71g/mol.
Therefore, we can write that-
Number of moles of chlorine = $\dfrac{213g}{71g/mol}=3\text{ }mol$
We can see from the above calculation that we have 3 moles of chlorine and 2 moles of calcium.
Now, the reaction given to us is: $Ca\left( s \right)+C{{l}_{2}}(g)\to CaC{{l}_{2}}\left( s \right)$
Here, calcium is used completely as we can see from the calculation that the number of moles of calcium present is less than that of chlorine.
Therefore, we can say that chlorine is present in excess of chlorine gas.
Now, the amount of chlorine gas present in excess = 3mol – 2mol = 1 mol.
So, we have 1mol excess chlorine gas.
Now we can convert the mole in grams to get the answer. We can do this conversion by multiplying the number of moles by the molar mass.
So, amount of excess chlorine gas = $1mol\times 71g/mol=71g$

Therefore, the correct answer is option (B) 71 g $C{{l}_{2}}$ excess.

Note: In the above question, calcium is the limiting reagent. In a chemical reaction the reactant which is totally consumed while the formation of the product is known as the limiting reagent. Calcium is the limiting agent because it is present in the lesser amount in this reaction. Limiting reagent is the substance that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reaction, since the reaction cannot continue without it.