People that are good at chem...PLEASE EXPLAIN!!!?

Question

A 0.250 L solution of 1.150 M hydrochloric acid is used to neutralize a 45.0 g sample of barium hydroxide in a coffe-cup calorimeter. Barium chloride, a chemical that is sometimes used to produce a bright green color in fireworks, is produced by the reaction. The temp. in the calorimeter was observed to rise from 20.0 C to 32.9 C during the course of the neutralization. Assume that the density of the hydrochloric acid solution is 1.00 g/mL and that the specific heat capacity of the system is the same as that for water, 4.18 J/(g*K). Assume that only water is available to absorb or release heat and that the reaction is at a constant pressure.
2HCl(aq)+Ba(OH)2(aq)-->BaCl2(aq)+2H2O(l)

1.What was the limiting reagent in the neutralization reaction? Provide support for your answer.

2. How many moles of excess reagent were left over?

3. If 27.3g of barium chloride were obtained after the neutralization was complete, what was the percent yield?

4. what is the enthalpy change for the system in the coffee-cup calorimeter?

PLEASE HELP ME! im so lost... :(

Answer 1

1) first calculate moles of each component. let's start wth hydrochloric acid, HCl

moles HCl = liters solution x moles per liter of solution
(note that molarity is moles per liter)

so moles HCl = 0.250 L x 1.150 moles / L = 0.288 Moles HCl

moles Ba(OH)2 = mass Ba(OH)2 / Molecular Weight Ba(OH)2
= 45.0 g / (171.3 g/mole) = 0.263 moles

notice both moles have three digits? 0.250 L has three significant digits.

your balanced equation says 2 moles HCl react with 1 mole Ba(OH)2 so that 0.288 moles HCl would react with 0.144 moles Ba(OH)2. Since you have more Ba(OH)2 moles than the 0.144 that HCl would react with, Ba(OH)2 is in excess and HCl is the limiting reagent.

2) 0.263 - 0.144 = 0.119 moles

3) % yield = actual yield / theoretical yield

so theoretically, 0.288 moles HCl would give 0.144 moles BaCl2 .. Again from the balanced equation that says 2 HCl + 1 Ba(OH)2 ----> 1 BaCl2

so actual moles = actual mass / molecular weight
= 27.3 g / (208.3 g / mole) = 0.131 moles

and % yield = 0.131 / 0.144 x 100% = 91.0%


4) given that water alone absorbed the heat, I would work this problem like this. First, .250 L of solution contains less water than 0.250 L of pure water (because you have HCl making up some of the solution), however, the HCl is reacting with Ba(OH)2 to make water. And since the density is the same and the heat capacity is the same for the solution relative to water, just assume the amount of water present is 0.250 liters.

So here we go...

change in enthalpy (call it dH) = m x cp x delta T (dT that is)
so
dH = .250 L x (1000 ml/ L) x (1 g / ml) x (4.18 J / (g °K) ) x {(32.9 +273.15)°K - (20+ 273.15)°K} = 13,500 J = 13.5 KJ

note the 273.15 is the conversion from C to K. but I could have left it off since delta °C = delta °K