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We wish to determine how many moles of [tex]$BaSO_4$[/tex] form when 50.0 mL of 0.250 M [tex]$Al_2(SO_4)_3$[/tex] reacts with excess [tex][tex]$Ba(NO_3)_2$[/tex][/tex].

[tex]\[
\begin{aligned}
3 Ba(NO_3)_2(aq) + Al_2(SO_4)_3(aq) & \rightarrow 3 BaSO_4(s) + 2 Al(NO_3)_3(aq)
\end{aligned}
\][/tex]

In the previous step, you determined

[tex]\[
0.0125 \text{ mol } Al_2(SO_4)_3 \text{ react. }
\][/tex]

How many moles of [tex]$BaSO_4$[/tex] form during the reaction?



Answer :

GinnyAnswer
To determine how many moles of [tex]\( \text{BaSO}_4 \)[/tex] form, let’s refer to the stoichiometry of the balanced chemical equation:

[tex]\[ 3 \text{Ba}(\text{NO}_3)_2(aq) + \text{Al}_2(\text{SO}_4)_3(aq) \rightarrow 3 \text{BaSO}_4(s) + 2 \text{Al}(\text{NO}_3)_3(aq) \][/tex]

From the balanced equation, we can see that 1 mole of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex] reacts to form 3 moles of [tex]\( \text{BaSO}_4 \)[/tex].

Given the previous calculation:

[tex]\[ 0.0125 \, \text{mol} \, \text{Al}_2(\text{SO}_4)_3 \text{ react} \][/tex]

Using the stoichiometric ratio from the balanced equation:

[tex]\[ \text{Moles of BaSO}_4 \text{ formed} = 3 \times \text{Moles of Al}_2(\text{SO}_4)_3 \][/tex]

Substituting the given moles of [tex]\( \text{Al}_2(\text{SO}_4)_3 \)[/tex]:

[tex]\[ \text{Moles of BaSO}_4 \text{ formed} = 3 \times 0.0125 \, \text{mol} \][/tex]

[tex]\[ \text{Moles of BaSO}_4 \text{ formed} = 0.0375 \, \text{mol} \][/tex]

Therefore, the number of moles of [tex]\( \text{BaSO}_4 \)[/tex] formed during the reaction is [tex]\( 0.0375 \, \text{mol} \)[/tex].

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