dry gaseous hcl contained in volume of 1L at STP is dissolved in 200ml of 0.5M barium hydroxide soltion. How many mol of 0.5M hcl would be needed to neutralise the resulting solution?
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
mols Ba(OH)2 = M x L = ?
mols HCl = twice that.
how many moles total does it take to neutralize the Ba(OH)2?
moles= 2*.2*.5=.2moles HCl
next, how much did was added as a gas?
moles= 1/22.4
How many moles do you need still?
.2-1/22.4
finally,
.2-1/22.4=.5*litersHCl
solve for the volume of .5M HCl.
To determine how many moles of 0.5M HCl are needed to neutralize the resulting solution, we need to first understand the reaction that occurs between HCl and Ba(OH)2. The balanced chemical equation for the reaction is as follows:
2HCl + Ba(OH)2 → BaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 1 mole of BaCl2 and 2 moles of water.
Now, let's calculate the number of moles of Ba(OH)2 present in the 200 mL solution of 0.5M Ba(OH)2:
Molarity (M) = Moles/Liter
Given:
Volume of Ba(OH)2 solution = 200 mL = 0.2 L
Molarity of Ba(OH)2 solution = 0.5 M
Moles of Ba(OH)2 = Molarity × Volume
= 0.5 mol/L × 0.2 L
= 0.1 mol
According to the stoichiometry of the balanced equation, 1 mole of Ba(OH)2 reacts with 2 moles of HCl. Therefore, 0.1 moles of Ba(OH)2 will react with 2 × 0.1 = 0.2 moles of HCl.
Hence, 0.2 moles of 0.5M HCl would be needed to neutralize the resulting solution.