how many milliliters of 6.00 M HCl are required to react with 60.0 ml of 1.96 M Al(OH)3?
Al(OH)3(s)+3HCl(aq)-->AlCl3(aq)+3H2)(aq)
Here is an example of a stoichiometry problem I've posted. To obtain moles, remember moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of milliliters of 6.00 M HCl required to react with 60.0 mL of 1.96 M Al(OH)3, we need to use the balanced chemical equation and stoichiometry.
Let's start by writing the balanced chemical equation:
2Al(OH)3(s) + 6HCl(aq) → 2AlCl3(aq) + 6H2O(l)
From the balanced equation, we can see that it takes 6 moles of HCl to react with 2 moles of Al(OH)3. This means that the molar ratio of HCl to Al(OH)3 is 6:2 or 3:1.
Now, let's calculate the number of moles of Al(OH)3 we have:
1.96 M = 1.96 moles/L
60.0 mL = 60.0/1000 L = 0.06 L
moles of Al(OH)3 = 1.96 moles/L × 0.06 L = 0.1176 moles
Since the stoichiometry ratio is 3:1 for HCl to Al(OH)3, we can determine the number of moles of HCl needed:
moles of HCl = 0.1176 moles of Al(OH)3 × (3 moles of HCl/1 mole of Al(OH)3) = 0.3528 moles
Now, we can calculate the volume of 6.00 M HCl needed:
6.00 M = 6.00 moles/L
moles of HCl = 0.3528 moles
volume of HCl = moles of HCl / Molarity = 0.3528 moles / 6.00 moles/L
volume of HCl = 0.0588 L = 58.8 mL
Therefore, 58.8 mL of 6.00 M HCl is required to react with 60.0 mL of 1.96 M Al(OH)3.