# How much Aluminum Chloride in grams is needed to completely precipitate 1.74 grams of K2SO4 (FW=174 g/mol)?

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1
Mar 5, 2018

Here it is: 0.89 grams

#### Explanation:

$3 {K}_{2} S {O}_{4} + 2 A l C {l}_{3} \to A {l}_{2} {\left(S {O}_{4}\right)}_{3} + 6 K C l$

is the balanced chemical reaction.

Now you have $\frac{1.74}{174} = 0.01$ moles of ${K}_{2} S {O}_{4}$

Therefore you need 0.0067 mole of $A l C {l}_{3}$

(MW of it is 133.5 grams). Therefore total amount needed to complete this reaction is $= 0.0067 \times 133.5 = 0.89$ grams.

You need 0.89 grams of $A l C {l}_{3}$ to precipitate 1.74 grams of ${K}_{2} S {O}_{4}$

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

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anor277 Share
Mar 5, 2018

The question is unsound...

#### Explanation:

We could change it to how much $\text{barium nitrate}$ is required to react with $1.74 \cdot g$ of $\text{potassium sulfate..??}$

And here, we would write the stoichiometric equation...

$B a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + {K}_{2} S {O}_{4} \left(a q\right) \rightarrow B a S {O}_{4} \left(s\right) \downarrow + 2 K N {O}_{3} \left(a q\right)$

And thus moles of potassium sulfate is equal to moles of barium nitrate.

$\text{Moles of potassium sulfate} = \frac{1.74 \cdot g}{174.26 \cdot g \cdot m o {l}^{-} 1} = 0.010 \cdot m o l$

And thus we require an equivalent quantity of barium nitrate...

$= 0.010 \cdot m o l \times 261.34 \cdot g \cdot m o {l}^{-} 1 = 2.61 \cdot g$...

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