Answer:
[tex]V=43.46mL[/tex]
Explanation:
Hello!
In this case, since the reaction between sulfuric acid and aluminum hydroxide is:
[tex]3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O[/tex]
Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:
[tex]n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4[/tex]
Then, given the molarity, it is possible to obtain the milliliters as follows:
[tex]V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL[/tex]
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