For the given reaction;
$$CaCO_{3}+2HCl\rightarrow CaCl_{2}+H_{2}O+CO_{2}$$
If 90 g $$CaCO_{3}$$ is added to 300 mL of HCI which contains 38.55% HCI by mass and has density 1.13 g $$mol^{-1}$$, then which of the following option is correct ?
Given molar mass of H, Cl, Ca and Oare 1, 35.5, 40 and 16 g $$mol^{-1}$$ respectively.

The reaction is: $$CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$$

Given:

• Mass of $$CaCO_{3}$$ = 90 g
• Volume of HCl solution = 300 mL
• Density of HCl solution = 1.13 g/mL (assuming typo in unit, should be g/mL instead of g mol^{-1})
• HCl solution contains 38.55% HCl by mass
• Molar masses: H = 1 g/mol, Cl = 35.5 g/mol, Ca = 40 g/mol, O = 16 g/mol

First, calculate the mass of the HCl solution:

Mass of HCl solution = density × volume = 1.13 g/mL × 300 mL = 339 g

Next, calculate the mass of HCl in the solution:

Mass of HCl = (38.55 / 100) × 339 g = 0.3855 × 339 g

Compute: 0.3855 × 300 = 115.65, 0.3855 × 39 = 15.0345, total = 115.65 + 15.0345 = 130.6845 g

So, mass of HCl = 130.6845 g

Now, find the molar masses:

Molar mass of $$CaCO_{3}$$ = Ca + C + 3O = 40 + 12 + 3×16 = 40 + 12 + 48 = 100 g/mol

Molar mass of HCl = H + Cl = 1 + 35.5 = 36.5 g/mol

Calculate moles of each reactant:

Moles of $$CaCO_{3}$$ = mass / molar mass = 90 g / 100 g/mol = 0.9 mol

Moles of HCl = mass / molar mass = 130.6845 g / 36.5 g/mol ≈ 3.5804 mol

(Calculation: 130.6845 ÷ 36.5 = 3.58039726... ≈ 3.5804 mol)

From the reaction stoichiometry, 1 mol of $$CaCO_{3}$$ requires 2 mol of HCl.

Moles of HCl required for 0.9 mol of $$CaCO_{3}$$ = 2 × 0.9 = 1.8 mol

Since available moles of HCl (3.5804 mol) are greater than required (1.8 mol), HCl is in excess, and $$CaCO_{3}$$ is the limiting reactant.

Thus, $$CaCO_{3}$$ will be completely consumed, and some HCl will remain unreacted.

Calculate the mass of HCl reacted:

Moles of HCl reacted = 1.8 mol

Mass of HCl reacted = moles × molar mass = 1.8 mol × 36.5 g/mol = 65.7 g

Mass of unreacted HCl = total mass of HCl - mass of HCl reacted = 130.6845 g - 65.7 g = 64.9845 g ≈ 64.98 g

Now, evaluate the options:

• Option A: 32.85 g of $$CaCO_{3}$$ remains unreacted. But $$CaCO_{3}$$ is completely consumed, so this is incorrect.
• Option B: 64.97 g of HCl remains unreacted. Our calculation gives 64.98 g, which is very close (likely a rounding difference).
• Option C: 97.30 g of HCl reacted. But we have 65.7 g reacted, so incorrect.
• Option D: 60.32 g of HCl remains unreacted. But we have approximately 64.98 g, so incorrect.

Thus, option B is correct.