Question

20ml of 0.1 M BaCl2 is mixed with 30ml o0f 0.2 M al2(SO4)3. What is the weight of BaSO4 formed?

Answer 1

The weight of BaSO_4 is 0.699 g

Given:

Molarity of BaCl_2 = 0.1 M

Volume of BaCl_2 = 20 ml

Molarity of $Al_2(SO_4)_3$ = 0.2 M

Volume = 30 ml

Solution:

The chemical reaction for this process is given below:

$3 \mathrm{BaCl}_{2}+\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \rightarrow 3 \mathrm{BaSO}_{4}+2 \mathrm{AlCl}_{3}$

From the balanced equation, we can see that 3 moles of $BaCl_2$ reacts with 1 mole of $Al_2(SO_4)_3$ to give 3 moles of $BaSO_4$

$\text{Number of moles} =\left(\frac{\text {Weight}}{\text {Molecular Weight}}\right) \times 1000$

We know that, molecular weight of $BaSO_4$ is 233 g/mol

$\text{Weight} =\left(\frac{(\text {Number of moles}) \times(\text {Molecular Weight})}{1000}\right)$

$=\frac{3 \times 233}{1000}$

$\Rightarrow \text { Weight of } B a S O_{4}=0.699 \ grams$

Answer 2

Answer:

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