Question

You will be observing a reaction in this lab. As with most reactions, this reaction occurs by the breaking of some bonds in the reactants and the formation of new bonds in the products. During the reaction, the gas that is produced will fizz away. After the reaction is complete, one of the products can be removed. Then you will determine the mass of the other product this will be your actual (or experimental) yield. You will compare this value to what you should have got (theoretical yield) to calculate your % yield. 1. Balance the following equation: NaHCO, + HCI Naci + H2O + CO2 2. List the reactants that will be put into the beaker: 3. List the products that will be formed in this reaction: 4. Which product will fizz away? 5. How can you get rid of another one of the products so that there is only one left? 6. How many grams of NaCl should be produced if you start with 1.0 g of sodium bicarbonate. This conversion will require three conversion factors. Start with 1.0 g NaHCO,. Then use a conversion factor that converts that to molos. Then use the balanced equation to convert from moles of NaHCO, to moles of NaCl. Then use a conversion factor to convert to grams. 7. This experiment will also make CO molecules which will float away into the atmosphere after fizzing out of the solution. How many molecules of CO2 will be made from 1.0 g sodium bicarbonate. This will also take three conversion factors. Start with 1.0 g NaHCO3. Then use a conversion factor that converts to moles. Then use the balanced equation to convert from moles of NaHCO, to moles of CO2. Then use one more conversion factor to convert from moles to molecules (think Avogadro). dicta

Answer 1

a) The reactants in the beaker are:

NaHCO3 and HCl.

b) The products formed are:

NaCl, H2O and CO2

c) The product that escapes from the beaker is CO2.

d) If the solution is heated to a little more than 100 ° C, the H2O evaporates and only NaCl remains.

e) Calculate the mass of NaCl to be formed:

m NaCl = 1 g NaHCO3 * (1 mol NaHCO3 / 84 g) * (1 mol NaCl / 1 mol NaHCO3) * (58.44 g NaCl / 1 mol) = 0.70 g

7) The CO2 molecules are calculated:

CO2 molecules = 1 g NaHCO3 * (1 mol NaHCO3 / 84 g) * (1 mol CO2 / 1 mol NaHCO3) * (6.02E23 molecules CO2 / 1 mol) = 7.17E21 molecules

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