Balanced Equation Lab Report

Mole of chlorine : 1.0217g - .221g - .3946 g = .4061 g of chlorine

3. Find the number of atoms of each of the substances involved in the reaction.

Aim: The aim of the lab “Chemical Equilibrium” is to observe the effects of changes in concentrations of products and reactants on the position of the equilibrium of given chemical reactions.

How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?

In your laboratory notebook sum these two reactions to find the stoichiometric factor that relates moles of

9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?

Once you have the mass of the gas, calculate the number of moles of CO2 that were produced in the chemical reaction between the vinegar and the 2 Alka-Seltzer® tablets?

Molarity for sol. # 2 = (0.0016 g/l)/(350.13 g/mol) = 4.6 × 10-6 moles/ l

Data Analysis: When weighing the mass of our product you get 2.13 grams. Also when you figure out the percent yield you get 93.4%.

= (Actual Yield / Theoretical Yield) * 100 % = (2.5707 g / 3.75 g) * 100 %

The goal of the stoichiometry lab was to use stoichiometry and percent yield to compare how much product was supposed to be produced with how much was actually produced by experimentation. In this lab, the reactants aluminium and copper (II) chloride were used to form aluminium chloride and copper.

This number means that the reaction is fairly product-favored as the number is larger than one. The actual value for the equilibrium constant of this reaction was within the range of 200-250. Our calculated value was fairly far off, with between 25-56% error. It is clear that the experimental data has error, as evidenced by the R2 value of .3092. It should be noted that if results from the fourth, fifth, and sixth addition of 1 mL of Fe(NO3)3 are omitted in Figure 1, the R2 jumps to .8451 and the equilibrium constant is 238.1, well within the desired

In this case, Hydrogen has 3 atoms present and there is only 1 atom present in nitrogen. 3. Multiply each element by the number of atoms present. N- 14.01 x 1 = 14.01 H- 1.01 x 3 = 3.03 4. Now, you can get the molar mass by adding each element's product. 14.01 + 3.03 = 17.04 5.

5. In reaction three, the number of moles of NaOH can be calculated from the concentration of the solution (1.0M = 1.0mole/L) and the volume used. The calculation is below. Enter the result into Data Table 2.

4g of Na CO x 1 mol of Na CO 4g of CaCl x 1mol of CaCl