Chemistry Help!?

Question

Here is the balanced equation:
NaHCO3 + HCl = NaCl + H2O + CO2
I started with 0.36 g of NaHCO3, and 5 mL of HCl. My product was 0.25 g of NaCl. The mole ratio of NaHCO3 to NaCl is 1:1. How do I calculate the Theoretical Yield of NaCl?

Answer 1

This equation is simple, just take NaHC03 and add HCI and you get NaCI, I dont know I give up!

Answer 2

Lancenigo di Villorba (TV), Italy

The reaction involved in the question is the following one :

NaHCO3(aq) + HCl(aq) ---> NaCl(aq) + H2O(aq) + CO2(g)

You wrote that the system spent 0.36 g of NaHCO3, e.g. a mass's datum equivalent to another one

(Soda's Moles) = (Soda's Mass) / MW = 0.36 / 84.00 =
= 4.286E-3 mol

You reported that the system given 0.25 g of NaCl, e.g. a mass's datum equivalent to another one

(Salt's Moles) = (Salt's Mass) / MW = 0.25 / 58.44 =
= 4.278E-3 mol

In the pathway passing from the SODA (e.g. NaHCO3) toward the SALT (e.g. NaCl) I highlight the SODIUM ION's CONNECTION existing between the two Chemical Compounds.
In the mathematical terms, I retrieve a Yield Factor as

100% * 4.277E-3 / 4.286E-3 = 99.81 mol%

I hope this helps you.

Answer 3

Convert the .36g to moles. Theoretically, you should have the same moles of NaCl on the other side because the equation says for 1 mole of NaHCO3 you get 1 of NaCl. This is your theoretical yield.

Figure out how many moles .25g of NaCl is (.25g * mole/g NaCl). Your % yield is:

actual moles NaCl/theoretical moles NaCl * 100%

Answer 4

Here you go. Go for it! It's your baby!http://classes.colgate.edu/pjue/chem263/yields.htm