If 200 mL of 3.70 M aqueous Ba(OH)2 and 980 mL of 0.837 M aqueous HClO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 3.70 M aqueous Ba(OH)2 remain? Round your answer to 3 significant figures.
Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)
answer = 89.2....
my calculations:
Ba(OH)2 = 0.74 su
HClO4 = 0.410 su
so... HClO4 is the limiting reagent
what do i do next?....
Good work so far.
So you have 0.740 moles Ba(OH)2
You have 0.410 moles HClO4 reacting
What's left is 0.740-0.410 = 0.330 moles Ba(OH)2.
M = moles/L
You know M and moles, solve for L and convert to mL, then round to three s.f.
To determine the answer, you need to calculate how much of the limiting reactant (HClO4) reacts with Ba(OH)2. This will allow you to find out how much Ba(OH)2 is consumed in the reaction.
First, calculate the number of moles of HClO4 using the given concentration and volume:
0.837 M * 0.98 L = 0.818 mol HClO4
According to the stoichiometry of the balanced equation, the ratio between HClO4 and Ba(OH)2 is 2:1. This means that for every 2 moles of HClO4, 1 mole of Ba(OH)2 reacts.
Next, use the stoichiometry to calculate the number of moles of Ba(OH)2 reacted:
0.818 mol HClO4 * (1 mol Ba(OH)2 / 2 mol HClO4) = 0.409 mol Ba(OH)2
Now, calculate the remaining moles of Ba(OH)2:
0.74 mol Ba(OH)2 - 0.409 mol Ba(OH)2 = 0.331 mol Ba(OH)2
Finally, convert the moles of Ba(OH)2 back to milliliters using the given molarity:
0.331 mol Ba(OH)2 * (1 L / 3.70 mol) * (1000 mL / 1 L) = 89.5 mL
So, approximately 89.5 mL of 3.70 M aqueous Ba(OH)2 will remain. Rounded to 3 significant figures, the answer is 89.2 mL.