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What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(OH) 3 (aq) Al 2 (SO 4 ) 3 (aq)+6 H 2 O

Sagot :

sebassandin

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]

Best regards!

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