Question

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

Answer 1

`V=43.46mL`

Step-by-step explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

`3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O`

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:

`n_(H_2SO_4)=3.209gAl(OH)_3*(1molAl(OH)_3)/(78.00gAl(OH)_3) *(3molH_2SO_4)/(2molAl(OH)_3) \\\\n_(H_2SO_4)=0.0617molH_2SO_4`

Then, given the molarity, it is possible to obtain the milliliters as follows:

`V=(n)/(M)=(0.0617mol)/(1.420mol/L)*(1000mL)/(1L)\\\\V=43.46mL`

Best regards!