We mixed HCl and NaOH into water and got this table, all for trial 1
Here is all of the info for trial 1:
trial 1:
volume of 2 M NaOH used: 50.0 mL
initial temperature of NaOH: 24.4 degrees C
volume of 2 M HCl used: 50.0 mL
initial temperature of HCl: 24.3 degrees C
final temperature reached: 36.1 degrees C
total mass (volume) of mixture: 100.0 g
temperature change, delta t: NAOH: 11.7 degrees C, HCl 11.8 degrees C
Heat flow, joules: 4932.0 J
Moles of water produced: ????
delta H (kj, mol water) ????
How do I find the moles of water produced and delta H (kj, mol water)?
Please help
To find the moles of water produced, you can use the concept of stoichiometry. The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> H2O + NaCl
From the equation, the stoichiometric ratio between HCl and water is 1:1. This means that for every 1 mole of HCl that reacts, 1 mole of water is produced.
Given that you used 50.0 mL of 2 M HCl, we need to determine the moles of HCl used.
To find the moles of HCl, you can use the formula:
moles = concentration (M) x volume (L)
First, convert the volume from mL to L:
50.0 mL = 50.0 mL x (1 L / 1000 mL) = 0.050 L
Now, calculate the moles of HCl:
moles HCl = 2 M x 0.050 L = 0.100 moles
Since the stoichiometric ratio is 1:1, the moles of water produced will also be 0.100 moles.
To find the enthalpy change (delta H) for the reaction (in kJ/mol of water), you can use the formula:
delta H = heat flow (J) / moles of water produced
From the given information, the heat flow is 4932.0 J and the moles of water produced is 0.100 moles.
delta H = 4932.0 J / 0.100 moles = 49320 J/mol
Finally, convert the answer to kJ/mol by dividing by 1000:
delta H = 49.32 kJ/mol (rounded to two decimal places)
Therefore, the moles of water produced is 0.100 moles and the enthalpy change (delta H) for the reaction is 49.32 kJ/mol.
To find the moles of water produced and delta H (kJ/mol water), you need to use the information provided and apply the principles of stoichiometry and the heat equation.
1. Calculate the moles of NaOH used:
To find the moles of NaOH used, you need to use the volume and concentration. In this case, the volume of 2 M NaOH used is given as 50.0 mL. Since the concentration is 2 M (moles per liter), you can convert the volume to liters by dividing it by 1000 mL/L:
Volume (L) = 50.0 mL / 1000 mL/L = 0.05 L
Now, you can calculate the moles of NaOH using the volume and concentration:
Moles of NaOH = Volume (L) × Concentration (M) = 0.05 L × 2 M = 0.10 moles
2. Calculate the moles of HCl used:
Using the same process, you can find the moles of HCl used. The volume of 2 M HCl used is also given as 50.0 mL.
Volume (L) = 50.0 mL / 1000 mL/L = 0.05 L
Moles of HCl = Volume (L) × Concentration (M) = 0.05 L × 2 M = 0.10 moles
3. Calculate the moles of water produced:
From the balanced chemical equation for the reaction between HCl and NaOH, we know that the stoichiometric ratio is 1:1. This means that for every mole of HCl reacted, one mole of water is produced.
As both the HCl and NaOH have reacted completely, the moles of water produced will be equal to the moles of HCl used:
Moles of water produced = Moles of HCl used = 0.10 moles
4. Calculate the delta H (kJ/mol water):
The heat flow in joules (4932.0 J) can be converted to kilojoules using the conversion factor of 1 kJ = 1000 J:
Heat flow (kJ) = 4932.0 J / 1000 = 4.932 kJ
Delta H is the heat flow divided by the moles of water produced:
Delta H (kJ/mol water) = Heat flow (kJ) / Moles of water produced
In this case:
Delta H (kJ/mol water) = 4.932 kJ / 0.10 moles = 49.32 kJ/mol water
Therefore, the moles of water produced are 0.10 moles, and the delta H (kJ/mol water) is 49.32 kJ/mol water.