chemistry!!! help me pleaseee!?

Question

How many milliliters of 20.9 M H2SO4 are required to react with 250. mL of 2.13 M Al(OH)3 if the products are aluminum sulfate and water?



helppp mee!!

Answer 1

set up the rxn equation and then use the equation MaVa/Ca = MbVb/Cb and plug in the numbers you have.

the M= molarity
the V= volume
the C= the stoichiometric coefficient

make sure you balance the rxn correctly too otherwise your "C" value will be off

the "a" stands for the acid (so you know which side the numbers go on- acid on one side, base on the other)
the "b" stands for the base

Answer 2

First, you have to write a balanced equation for the reaction. In this case, it will be:

3 H2SO4 + 2 Al(OH)3 --> Al2(SO4)3 + 6 H2O

Now, start with the Al(OH)3 and calculate how many moles of that you have be multiplying the molarity times the volume (in Liters). Next, use the coefficients of the equation to convert moles of Al(OH)3 into moles of H2SO4. To do that, you'll multiply the moles of Al(OH)3 by 3/2. Finally, divide that by the concentration of the H2SO4 solution to convert moles of sulfuric acid into Liters of H2SO4 solution. Lastly, multiply that by 1000 to get mL of solution.