Consider the reaction: 2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ

Question

a) Calculate the heat (in kJ) released (make sure to use the correct sign in your answer) when 106.7 mL of 0.365 M HCl is mixed with excess Ba(OH)₂.

b) How many grams of barium hydroxide are required to produce enough energy to heat a 509.8 g of sample of water from 22.3°C to 96.6°C?

Answer 1

a)

0.365 M HCl = 0.365 moles/ 1 L

we have 106.6 mL, which is 0.1066 L

(0.365 M HCl)(0.1066 L) = 0.039 moles HCl

Using our Balanced Chemical Equation

2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = -118 kJ

(0.039 mole HCl)(-118kJ / 2 HCl) = -2.301 kJ

b) How many grams of barium hydroxide are required to produce enough energy to heat a 509.8 g of sample of water from 22.3°C to 96.6°C?

This one is much more difficult.

First we need to find the amount of energy needed to heat the water

q=mCΔT

q=(509.8 g)(4.184 J/g°C)((96.6°C-22.3°C))

q=(509.8 g)(4.184 J/g°C)(74.3°C)

q=158,482 J

q= 158.5 kJ

This is how much energy would be needed to heat 509.8 g of water from 22.3 °C to 96.6 °C

Our balanced chemical equation tell us that for every one mole of Ba(OH)₂ we will produce 118 kJ of energy (-118 kJ).

(1 mol Ba(OH)₂)/(-118 kJ) is our conversion factor

(-158.5 kJ)(1 mol Ba(OH)₂)/(-118 kJ) = 1.34 moles Ba(OH)₂ needed

The question wants grams Ba(OH)₂

to convert, we will use the molar mass of Ba(OH)₂

171.34 g/mol

1.34 mol)(171.34g/mol) = 229.6 grams Ba(OH)₂ needed