Question-132337

Question Number 132337 by I want to learn more last updated on 13/Feb/21

Answered by physicstutes last updated on 14/Feb/21

First we determine the limiting reagent :

mass of {BaCl}_{2} = 40.8 g

\Rightarrow number of moles = \frac{40.8}{(137.3 + 2 \times 35.5)} mol = 0.196 mol

mass of K_{2} {SO}_{4} = 50.7 g

\Rightarrow number of moles = \frac{50.7}{(2 \times 39.0 + 32.0 + 4 \times 16)} mol = 0.291 mol

hence {BaCl}_{2} is the limiting reagent and will determine the end of the reaction

now we do a gram to gram conversion .

gram \to number of moles \to mole ratio \to gram .

\frac{0.196 mol {BaCl}_{2}}{1} \times \frac{2 mol KCl}{1 mol {BaCl}_{2}} \times \frac{74.5 g KCl}{1 mol KCl} = 29.2 g KCl

Commented by I want to learn more last updated on 23/Feb/21

I appreciate sir . Thanks

Commented by Tawa11 last updated on 14/Sep/21

nice

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