Partial Pressure and Initial D Equilibrium

For the reaction_, determine how many moles of chlorine Cl2 would be needed to react with

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.

Rate= k [I-]1[H2O2]. (2.13675*10-5 ) = k [0.015] [0.015] then solve for k. For this trial, k=0.09497.

3. You collect 14.2 liters of CO gas from an experiment. How many molecules would be in the sample?

(0.074 mol HCl x 1 mol NaOH) / 1 mol HCl = 0.074 mol NaOH

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

A 300.0-mL saturated solution of copper(II) peroidate (Cu(IO4)2) contains 0.38 grams of dissolved salt. Determine the Ksp.

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

Finally, we measured the mass of the remaining copper in the test tube. This lead to the discovery that the mass of the copper in relation to our original 1.08 grams of copper oxide was 0.96 grams. From this data, we could find that the measurement of oxygen in our amount of copper oxide was 0.12 grams. With these significant weights, we were then able to convert those results to the amount of moles each element had in contribution to our total quantity of copper oxide. Specifically, we found that there was 0.015 moles of copper in our copper oxide, and 0.0075 moles of oxygen. So, that made the calculated ratio between moles of copper and moles of oxygen to be 2.0, or a 2:1 ratio of copper to oxygen. Thus, the formula of the red copper oxide can be expressed as . The lab worked the way it did primarily because of the chemical characteristics of copper oxide. Since the oxygen could separate from the copper when presented with a heat source, it allowed for the amount of each element in our quantity of copper oxide to be measured and then calculated into a ratio, which determined the formula for the

Let’s start with HNO3: m= 725g, M=63.02g/mol, therefore n= 725g/63.02g/mol=11.5mol ("NITRIC ACID"). Now C7H8: m=750g, M= 92.05g/mol, thus n=750g/92.05g/mol=8.15mol ("Toluene"). Afterwards, using the equation 3HNO3(aq) + C7H8(s) → C7H5N3O6(s) we can determine the ratio between C7H8 and HNO3, which is 1:3, therefore HNO3 is the limiting reactant. To ding the number of moles there are in TNT, we will have to multiply the moles of the limiting reactant by the ratio of TNT and HNO3: 11.5mol of HNO3 x 1mol of C7H5N3O6/ 3 mol of HNO3= 3.83mol of C7H5N3O6. Afterwards we use the formula of n=m/M and m=

formula Kw[Fex(C2O4)y]·zH2O. The variables x, y, and z were determined through the duration of the

The literature value for the enthalpy change of the last reaction which was provided by our teacher is -97 kJ mol-1. Agreeing this value, our result can be considered accurate.

6 x C-H (412) + 2 x C-O (360) + 3 x O-H (463) + 3 x O=O (498)  4 x C=O (805) + 8 x H-O (463)

In this experiment, a redox reaction occurred. An oxidation-reduction reaction (redox) is a type of chemical reaction that involves a transfer of electrons between two solutions. The chemical being oxidized is losing electrons and the chemical being reduced is gaining electrons. In this case KMnO4 is losing electrons and Oxalic Acid is gaining electrons. KMnO4 can titrate a or reduce Oxalic acid. Titration is the technique used to find the unknown concentration of one solution based on the concentration of a known solution. In equation 2, the molar relationship between the 2 is shown, it is 2 KMnO4 to 5 Oxalic Acid. The molar ratio relationship is useful because it shows how much of a certain product is needed to help a reaction occur and which chemical is limiting reagent. The experiment was started by preparing a titrating strength KMnO4 solution from stock to a less concentrated KMnO4. Equation one shows how this was done. The KMnO4 needed to be diluted, if it had not been diluted, then it would be way too hard to get an accurate reading of the Oxalic Acid used. Without being diluted the Oxalic Acid would be strong. The next step was to standardize the KMnO4 solution. It was calculated that 37.5 mL of Oxalic Acid could titrate 15 mL of KMnO4. To determine the exact molarity of a solution a standardization needs to happen. In this experiment the standardization found how much Oxalic acid was needed. Through the controlled variable, the fact that 37.5 Oxalic acid could be titrated by 15 mL of KMnO4, the percent of the

Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to