a. if you have spilled 6.3 mol of hcl and put 2.8 mol of ca(oh)₂ on it, which substance is the limiting…

Explanation:

Step1: Write the balanced chemical equation

$Ca(OH)_2 + 2HCl\rightarrow CaCl_2+2H_2O$

Step2: Determine the mole - ratio

The mole - ratio of $Ca(OH)2$ to $HCl$ is $n{Ca(OH)2}:n{HCl}=1:2$.

Step3: Calculate the moles of $Ca(OH)_2$ needed to react with given $HCl$

Given $n_{HCl} = 6.3$ mol. According to the mole - ratio, the moles of $Ca(OH)2$ needed, $n{Ca(OH)2\ required}=\frac{n{HCl}}{2}=\frac{6.3}{2}=3.15$ mol.

Step4: Compare with the available moles of $Ca(OH)_2$

We have $n_{Ca(OH)_2\ available}=2.8$ mol. Since $2.8\ mol<3.15\ mol$, $Ca(OH)_2$ will be completely consumed first.

Answer:

$Ca(OH)_2$