Consider the following reaction

Ca + 2HCl → CaCl₂ + H₂

We have 14 g Ca reacts with excess of HCl. Choose the incorrect option.

To determine the incorrect option, let us analyze the stoichiometry of the given chemical reaction step-by-step.

The balanced chemical equation is:
$\mathrm{Ca}+2\mathrm{HCl}\to {\mathrm{CaCl}}_2+H_2$

First, we find the molar masses of the substances involved:
- Molar mass of Calcium ($\mathrm{Ca}$) ≈ 40 g/mol
- Molar mass of Calcium chloride (${\mathrm{CaCl}}_2$) ≈ 40 + 2 × 35.5 = 111 g/mol
- Molar volume of an ideal gas at STP ≈ 22.4 L/mol

Given mass of Calcium ($\mathrm{Ca}$) = 14 g.
Now, calculate the number of moles of Calcium reacting:
$\mathrm{Moles\; of\; Ca}=\frac{14\text{g}}{40\text{g/mol}}=0.35\text{mol}$

From the balanced chemical equation, 1 mole of $\mathrm{Ca}$ produces 1 mole of $H_2$ and 1 mole of ${\mathrm{CaCl}}_2$.
Therefore, the moles of products formed are:
- Moles of $H_2$ produced = 0.35 mol (This matches the second option, making it correct).
- Moles of ${\mathrm{CaCl}}_2$ produced = 0.35 mol

Let us calculate the volume of $H_2$ produced at STP:
$\mathrm{Volume\; of}{H}_2=0.35\text{mol}\times 22.4\text{L/mol}=7.84\text{L}$
(This matches the third option, making it correct).

Let us calculate the mass of ${\mathrm{CaCl}}_2$ produced:
$\mathrm{Mass\; of}{\mathrm{CaCl}}_2=0.35\text{mol}\times 111\text{g/mol}=38.85\text{g}$
(This matches the first option, making it correct).

Comparing these calculations with the fourth option, which states that the mass of ${\mathrm{CaCl}}_2$ produced is 3.885 g, we see that this value is incorrect. Thus, the incorrect option is: "Mass of CaCl2 produced is 3.885 g".