Question

\\\text{ba(oh)}_2 + 2\text{hbr} \
ightarrow \text{babr}_2 + 2\text{h}_2\text{o}\\

find the limiting reactant if \\(5.0\text{g}\\) of \\(\text{ba(oh)}_2\\) and \\(25.0\text{g}\\) of \\(\text{hbr}\\) are used in the reaction.

the limiting reactant is \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_

how many grams is the excess reactant in excess?

Explanation:

Calculate moles of each reactant

Using the Stoichiometry knowledge point

$$
LATEXBLOCK0
$$

Determine the limiting reactant

We compare the mole ratio of the reactants to the balanced equation:

$$ \text{Ba(OH)}_2 + 2\text{HBr} ightarrow \text{BaBr}_2 + 2\text{H2O} $$

Each mole of \(\text{Ba(OH)}_2\) requires 2 moles of \(\text{HBr}\).

$$ n_{\text{HBr, required}} = 2 \times n_{\text{Ba(OH)}_2} = 2 \times 0.02918\text{ mol} = 0.05836\text{ mol} $$

Since the available \(n_{\text{HBr}} = 0.3090\text{ mol}\) is greater than \(0.05836\text{ mol}\), \(\text{HBr}\) is in excess.
Therefore, \(\text{Ba(OH)}_2\) is the limiting reactant.

Calculate the mass of excess reactant remaining

We find the moles and mass of \(\text{HBr}\) that remain unreacted:

$$
LATEXBLOCK1
$$

Rounding to significant figures (two significant figures based on \(5.0\text{ g}\)):

$$ \text{Mass of excess HBr} \approx 20\text{ g} $$

Answer:

Question 1

The limiting reactant is \(\text{Ba(OH)}_2\)

Question 2

\(20\text{ g}\)