Stoichiometry Lab Report
The decomposition of sodium carbonate is definitely, Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with, when I balanced out the equation, I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g
NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence, I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards, I divided 339.1584 by 84.007 and got 4.04. Then, I divided 4.04 by 2 and got 2.02. In order to get my percent yield I divided 2.02 by 3.2 plus multiplied that by 100 getting 62.8. Now, when I had started of the experiment, I had
NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence, I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards, I divided 339.1584 by 84.007 and got 4.04. Then, I divided 4.04 by 2 and got 2.02. In order to get my percent yield I divided 2.02 by 3.2 plus multiplied that by 100 getting 62.8. Now, when I had started of the experiment, I had