1.00 g of sodium is added to a large excess of water under conditions of constant pressure (1.00 atm) and constant temperature (25 °C ). The following reaction occurs:
2Na (s) +2H2O (l)------> 2NaOH (aq) +H2(g)
Delta H°rxn =-369 kJ/mol
Calculate the heat released after 1.00 g of Na(s) fully reacted. Atomic weight of Na is 23.0 g/mol
Could you please tell me how to do this! Thank you!
369 kJ for 2*23 = 46 g. How much for 1.00 g? 369 x 1/46 = ?
I got an answer of 8.02...is this right? :)
No, but 8.02 kJ is. :-)
Ohhh yeahh haha thanks DrBob222! I really appreciate it! :D
Certainly! To calculate the heat released after 1.00 g of Na(s) fully reacts, we need to determine the number of moles of Na(s) that reacted and then use the molar enthalpy of reaction to calculate the heat released.
Here are the steps to solve the problem:
Step 1: Determine the number of moles of Na(s):
Using the atomic weight of Na (23.0 g/mol), we can calculate the number of moles by dividing the mass of Na(s) by its molar mass.
Number of moles = mass / molar mass
Number of moles = 1.00 g / 23.0 g/mol
Step 2: Calculate the heat released:
The molar enthalpy of reaction (ΔH°rxn) is given as -369 kJ/mol, which means that for every 2 moles of Na(s) reacted, -369 kJ of heat is released.
Therefore, for the number of moles calculated in step 1, we can set up a proportion to find the heat released.
Let's call the heat released as Q.
Number of moles of Na(s) / 2 moles = Q (kJ) / -369 kJ
Solving for Q:
Q (kJ) = (Number of moles of Na(s) / 2 moles) * -369 kJ
Plug in the number of moles of Na(s) calculated in Step 1 into the equation above to find the heat released.
I'll take care of the calculations for you.