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
3.2g of sodium bicarbonate. In the end, however, I was left with only 2g. Basically, when I divided 2 and 3.2 plus multiplied that by 100, I got 62.5g. Despite that fact, the calculations between the two were off by 0.3g, they are very similar. Additionally, none of the other problem’s percentages were even close enough to this. For further detail, ss sodium carbonate burns burns a lot of carbon dioxide is produced in the process. Nevertheless, a substance such as oxygen are slowly eliminated as well from the surrounding atmosphere. Production of products such as sodium carbonate and little amounts of water are likely to occur during the process. So, I can conclude that NaHCO3→ Na2CO3(s)+ CO2(g)+ H2O(g).
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
3.2g of sodium bicarbonate. In the end, however, I was left with only 2g. Basically, when I divided 2 and 3.2 plus multiplied that by 100, I got 62.5g. Despite that fact, the calculations between the two were off by 0.3g, they are very similar. Additionally, none of the other problem’s percentages were even close enough to this. For further detail, ss sodium carbonate burns burns a lot of carbon dioxide is produced in the process. Nevertheless, a substance such as oxygen are slowly eliminated as well from the surrounding atmosphere. Production of products such as sodium carbonate and little amounts of water are likely to occur during the process. So, I can conclude that NaHCO3→ Na2CO3(s)+ CO2(g)+ H2O(g).