Question

If 200 mL of N/10 HCl were added to 1g calcium carbonate what would remain after the reaction?
A. $CaC{{O}_{3}}$
B. HCl
C. Neither of the two
D. Parts of both

Answer 1

Write the reaction of calcium carbonate and HCl. When Calcium carbonate $CaC{{O}_{3}}$ reacts with HCl it gives Calcium chloride and some by products. Molecular mass of calcium carbonate is 100 g/mol. To convert normality into molarity, multiply normality with the n factor of the given element.

Complete step by step solution:
As you have learned about the concentration term in your chemistry lessons and how one concentration term is related with another.
From the question we have seen that concentration of HCl is given as N/10 and volume is given as 200ml. Here we have reacted with one gram of calcium carbonate with HCl.
So, the reaction between HCl and $CaC{{O}_{3}}$ is written as,

$CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O$
Here one mole of $CaC{{O}_{3}}$ reacts with two mole of HCl.
In this question we have to find the mass of the reactant to check what will be left after the reaction because the mass of calcium carbonate is given as 1g and if the mass of reactant will also be 1 g then neither HCl nor calcium carbonate will remain after the reaction.
Let us check what will be the mass of reactant,
Volume of HCl is given= 200ml= 200/1000 L
Normality of HCl =0.1
Now change the normality into molarity so we can find the moles because here volume is given,
So, Normality= molarity × n –factor
Molarity=$\dfrac{normality}{n-factor}$
For acids n- factor will be the no of hydrogen atom present in it. Thus for HCl the n-factor will be one
Thus molarity =$\dfrac{0.1}{1}=0.1$
And molar mass of $CaC{{O}_{3}}$ = 100g/mol
As we know that, Molarity= $\dfrac{no.\,of\,moles}{volume\,in\,L}$
Thus, no of moles of HCl= molarity × volume in L
No put the values in the formula we will get,
No. of moles = $0.1\times \dfrac{200}{1000}$
Here, calcium carbonate reacts with two mole of HCl,
Thus, no. of moles = $0.1\times \dfrac{200}{1000}\times \dfrac{1}{2}$= 0.01 moles.
Now to calculate the mass of reactant we have the formula as,
Moles= given mass/ molar mass
Mass of reactant = moles × molar mass = 0.01 × 100 = 1g
So, both calcium carbonate and HCl react completely during the reaction and nothing is left.

Thus the correct option will be (C).

Note: To react with one mole of calcium carbonate two moles of HCl is required. And if in the question normality and volume is given then you have to convert normality into molarity because normality has no connection with volume.