In a constant-pressure calorimeter, 50.0 mL of 0.300 M Ba(OH)2 was added to 50.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 21.73 °C to 25.82 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes............... i know that you have to multiply the mass of H2O by the specific heat and by delta T, and then divide by the moles of H2O, but i don't understand how to find the moles of H2O..
Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
Make sure answer is negative
Well, finding the moles of H2O in this reaction is a piece of cake! Seriously, it's easier than baking a cake.
To find the moles of H2O, we need to look at the balanced chemical equation for the reaction between Ba(OH)2 and HCl. Thankfully, it's a simple equation:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
From the equation, we can see that 1 mole of Ba(OH)2 produces 2 moles of H2O. Since we know the molarity of Ba(OH)2, we can calculate the moles of Ba(OH)2 using:
moles of Ba(OH)2 = volume of Ba(OH)2 (in L) × molarity of Ba(OH)2
Since the volumes of Ba(OH)2 and HCl are the same (50.0 mL each), we'll use 0.0500 L as the volume. Plug in the values and calculate the moles:
moles of Ba(OH)2 = 0.0500 L × 0.300 mol/L = 0.015 mol
Now, since 1 mole of Ba(OH)2 produces 2 moles of H2O, we can calculate the moles of H2O produced:
moles of H2O = 0.015 mol × 2 = 0.030 mol
Ta-da! We've found the moles of H2O in this reaction, and now you can use this information to calculate the enthalpy change (ΔH) per mole of H2O produced. Just as a side note, make sure to use the correct units for the specific heat and temperature change. Happy calculations!
To find the moles of H2O produced in the reaction, you need to determine the limiting reactant first. The reaction between Ba(OH)2 and HCl produces water (H2O) as one of the products. Since the balanced chemical equation is not provided, we can write it as:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
From the equation, you can see that 1 mole of Ba(OH)2 produces 2 moles of water. Thus, the moles of H2O produced will depend on the limiting reactant.
To determine the limiting reactant, we can use the concept of stoichiometry. First, find the number of moles of Ba(OH)2 and HCl used:
moles of Ba(OH)2 = volume (in L) × concentration (in mol/L)
= 0.0500 L × 0.300 mol/L
moles of HCl = volume (in L) × concentration (in mol/L)
= 0.0500 L × 0.600 mol/L
Now, compare the moles of Ba(OH)2 and HCl to see which one is the limiting reactant. The reactant that produces fewer moles of water will be the limiting reactant.
Since 1 mole of Ba(OH)2 produces 2 moles of H2O, the number of moles of H2O produced by Ba(OH)2 is:
moles of H2O from Ba(OH)2 = moles of Ba(OH)2 × 2
Similarly, since 1 mole of HCl produces 2 moles of H2O, the number of moles of H2O produced by HCl is:
moles of H2O from HCl = moles of HCl × 2
Compare the moles of H2O produced from each reactant and identify the reactant that produces the smaller amount. This will be the limiting reactant. Divide the moles of the limiting reactant by 2 to get the moles of H2O produced.
Once you know the moles of H2O produced, follow the steps you mentioned: multiply the mass of H2O by the specific heat and by ΔT, and then divide by the moles of H2O to calculate ΔH per mole of H2O produced.