chem 1 lab mole ratio in chemical reactions assignment

Question Description

I did the lab and took all the data that is attached. I need help with the post lab questions. attaches is the introduction of the lab, procedure, pre lab questions and answers, lab data, and post lab questions. Thank you!!

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Running head: MOLES RATIO IN CHEMICAL REACTION Moles Ratio in Chemical Reaction Course’s Name Student’s Name Professor’s Name Institution Due Date 1 Running head: MOLES RATIO IN CHEMICAL REACTION 2 Moles Ratio in Chemical Reaction Introductory Statement The purposes of this experiment were to determine the coefficient of the given reactions by using the data that records the temperature of the reaction and to identify which reactants were limiting and excess by the mean of graphical analysis. The theories used in this laboratory experiment were that the acid reacts with base to form the corresponding salt, and precipitation could be formed by letting two soluble salts react with each other; in addition, the formula that relates the number of moles, the molarity, and the volume was given as moles = molarity (moles/L) x volume (L). In terms of the procedure, the experiment was performed on LabQuest, where a temperature probe was connected to do the data collection for the temperature for all 8 trials where the volume of acid was kept constant but the volume of base and distilled water were varied. Several steps was done in LabQuest to record the lowest and highest temperatures, which would further be used in the graphical analysis that would determine the moles ratio between reactants as well as which reactants are limiting and excessive. Answer to Pre-lab Questions / Activities Question 1: Throughout the experiment, the reactant sulfuric acid ( H 2 SO4 ) was held constant in the first reaction where it reacts with NaOH to form the salt, and the reactant sodium phosphate ( Na3 PO4 ) was also held constant in the second reaction where it reacts with BaCl2 to form precipitation. These two reactants were held constant because, in each of the two reactions, we vary the other reactant and measure the heat released by the chemical reaction in all 8 trials. It is also accounted that in the second reaction, the Running head: MOLES RATIO IN CHEMICAL REACTION 3 amount of precipitation BaCl2 formed is directly proportionally to the amount of heat released that we measured in the reaction. Next, we can analyze the change in temperature graphically in order to determine the mole ratio of the reactions. That is why it is necessary to hold one reactant constant and vary the other constant in each of the 2 chemical reactions. The total volumes we used throughout the experiment were 125 mL of sodium hydroxide ( NaOH ) solution, 100 mL of sulfuric acid ( H 2 SO4 ) solution, and 150 mL of distilled water, and the above amounts were given in the procedure section and later on were divided into all 8 trials or reactions. If the actual concentration of the solution provided in the laboratory experiment was 1.0M, the number of moles of the reactant H 2 SO4 which is kept constant was calculated by:  1L  moles of H 2 SO4 = (1.0 M )(100 mL )    1000 mL   moles of H 2 SO4 = 0.1 moles Hence, the total number of moles H 2 SO4 in the initial solution which was further divided into 8 reactions is calculated to be 0.1 moles. Question 2: For the region in which the moles of barium chloride ( BaCl2 ) was greater than or equal to 0.003 mol, BaCl2 was limiting reagent. That was because when the moles of BaCl2 was greater than or equal to 0.003 mol, the data showed a horizontal line which implies that the amount of precipitation BaSO4 ( s ) did not increase any more. It means that the number of moles of the precipitation BaSO4 ( s ) was constant after the moles of BaCl2 Running head: MOLES RATIO IN CHEMICAL REACTION 4 exceeding 0.003 mol, which further implies that the moles of BaCl2 reacted to form the precipitation was not changed. Therefore, BaCl2 was the limiting reagent. A molecular-view of this point in the reaction can be drawn as follow: The above molecular-view of the reactants only illustrated that BaCl2 was the limiting reagent. Question 3: 3) The given chemical equation can be balanced as follow: 2 NaOH ( aq ) + H 2 SO4 ( aq ) → Na2 SO4 ( aq ) + 2H 2O (l ) Question 4: Based on the data given in the procedure section, the required table can be completed as follow: Moles NaOH Moles H2SO4 NaOH: H2SO4 Moles Na2SO4 Limiting Reactant Reactant Ratio Reactant Reactant Reaction 1 0.0025 0.01 0.25 0.00125 NaOH Reaction 2 0.005 0.01 0.5 0.0025 NaOH Reaction 3 0.0075 0.01 0.75 0.01 H2SO4 Reaction 4 0.01 0.01 1 0.01 H2SO4 Reaction 5 0.015 0.01 1.5 0.01 H2SO4 Running head: MOLES RATIO IN CHEMICAL REACTION 5 Reaction 6 0.02 0.01 2 0.01 H2SO4 Reaction 7 0.025 0.01 2.5 0.01 H2SO4 Reaction 8 0.03 0.01 3 0.01 H2SO4 ...
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