Mole and Mass Relationship in a Chemical Reaction: A Chemistry Experiment, Assignments of Chemistry

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Mole and Mass Relationship in a Chemical Reaction

General Chemistry I

Objective: In this experiment, we will predict a balanced equation for the reaction taking place, react to a known mass of NaHCO 3 with excess HCl, calculate the mole ratio between NaHCO 3 and NaCl., and determine whether your results support the Law of Conservation of Mass Results and Discussions: Table 1: shows the masses of measurements taken Mass (g) Calculations and observations Mass of glassware and watch glass 44.1428 g Mass of 𝑁𝑎𝐻𝐶𝑂 used 3 2.52 g 𝑁𝑎𝐻𝐶𝑂 is a white solid. When HCl was 3 added gas was released. Mass of glassware and watch glass & product 45.7668 g Mass of product 1.6240 g 45.7768g - 44.1428 g Post-lab Questions:

1. 0.0300 moles used in this reaction. 𝑁𝑎𝐻𝐶𝑂 3 𝑢𝑠𝑒𝑑 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑓𝑜𝑟 𝑁𝑎𝐻𝐶𝑂 3 =^ # 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠^ 2.52 𝑔 84.01 𝑔 =
2. NaCl→ 23 g + 35. 45 g = 58.45 g NaCl has the same number of moles as 𝑁𝑎𝐻𝐶𝑂 3 58.45 g x 0.0300 mol = 1.75 g We can produce 1.75g of moles of NaCl.
3. 1. 0300 𝑚𝑜𝑙 𝑥 (6. 023 𝑥 23 ) = 1. 80 𝑥 10 22
   2. 80 𝑥 10 moles of we would produce. 22 𝐻 2

a. An excess amount of HCl was added to make sure there were enough moles of HCl. 2