plz can anyone read my lab report n tell me if it´s correct?my whole chimestry course depends on it :( plzzz?
Heating of bicarbonate
The purpose of this lab is for us to find out which reactions are correct when sodium bicarbonate heats .
In this lab , we need to know how to find molar masses and convert it into molecules. We also need to know how to write a balanced chemical equation .
Amount of substance formula n = m / M,
Lot formula m = n * M ,
Molar mass formula M = n / m
Possible chemical reactions:
Proposal 1: NaHCO3 -> NaOH ( s ) + CO2 ( g )
Proposal 2: NaHCO3 -> Na2O ( s ) + H20 ( g) + CO2 ( g )
Proposal 3: NaHCO3 -> Na2CO3 ( s ) + H2O ( g) + CO2 ( g )
crucible , crucible tongs , wave, Burner , Spoon , tripod .
Chemicals : 1.50 g NaHCO3 ( s ) .
The lab began with the reaction formulas balanced and then weighed empty crucible to know how much it weighs without sodium bicarbonate. Afterwards added 1.50 g of sodium bicarbonate ( NaHCO3) in the crucible . Thereafter treforten for holding crucible of burners. Moreover, with the help of the wave was measured measured the mass of the product. How would it know which of the reactions were in the best, and it did using to calculate the molar mass and mass.
Balancing the reactions are:
1. NaHCO3 -> NaOH ( s ) + CO2 ( g )
NaHCO3 -> NaOH ( s ) + CO2 ( g )
2nd NaHCO3 -> Na2O ( s ) + H2O ( g) + CO2 ( g )
2NaHCO3 -> Na2O ( s ) + H2O ( g ) +2 CO2 ( g )
3rd NaHCO3 -> Na2CO3 ( s ) + H2O ( g) + CO2 ( g )
2NaHCO3 -> Na2CO3 ( s ) + H2O ( g) + CO2 ( g )
M (NaHCO3 ) : Na = 22.1 g, H = 1.01 g , C = 12.0 g , O = 16.0 g * 3 = 83.11 g / mol
m (NaHCO3 ): 1.50 g
n (NaHCO3 ): 1.50 g / 83.11 g / mol = 0.01804 = about 0.02 moles
After the reaction mass was : 0.6 g
1. NaHCO3 -> NaOH ( s ) + CO2 ( g )
M (NaOH ) = Na = 22.1 g + O = 16.0 g + H = 1.01 g = 39.11 g / mol
m (NaOH ) = 0.6 g
n = 0.6 g / 39.11 g / mol = 0.015341 = about 0.02 moles
Amount ( mol) is equal on both sides .
2nd 2NaHCO3 -> Na2O ( s ) + H2O ( g ) +2 CO2 ( g )
M ( Na2O ) = 22.1 g * 2 + O = 16.0 g = 60.2 g / mol
m ( Na2O ) = 0.6g
n ( Na2O ) = 0.6 g / 60.2 g / mol = 0.009966 = approximately 0.011 mol
number of moles is double led so great on the left side . In order for us to know if it is the right reaction , so we need to multiply by (2 ) to get the same number of moles on both sides -> n ( Na2O ) = 0.009966 * 2 = 0.019932 = approximately 0.021 mol
3rd 2NaHCO3 -> Na2CO3 ( s ) + H2O ( g) + CO2 ( g )
M (Na2CO3 ) = 22.1 * 2 + 1.01 + 16.0 * 3 = 93.2 g / mol
m (Na2CO3 ) = 0.6g
n (Na2CO3 ) = 0.6 g / 93.2 g / mol = 0.0643777 = approximately 0.0064 moles
The number of moles is double led so great on the left side . Then we multiply by (2 ) to get the same number of moles of both sides.
n (Na2CO3 ) = 0.0643777 * 2 = 0.0128755 = about 0.013 moles
Calculations of the amount of substance has been shown that heating of potassium hydrogen carbonate takes place according to reaction formulas 3rd
When sodium bicarbonate is heated , there was formed sodium carbonate, water and carbon dioxide. After the heating, the less mass as water and carbon dioxide is released , which means that they changed their aggregationsformler and became gases. The net amount calculated by the mass was already known and molar mass could be calculated, then it could take to be sorted out, please calculate the amount of substance for both sides and see which one has the most equal amount of substance with potassium bicarbonate. When the calculations were made , we did so with the form = n = m / M. This formula is used to calculate the amount of substance . In reactions 2-3 so had reactants twice more moles than the product because it was necessary to multiply the amount of substance by (2 ) to , know about substance quantities are equal on both sides. The heating of sodium bicarbonate takes place according to reaction # 3 to water, carbon dioxide and sodium carbonate is formed and for amount of substance is the most similar on both sides of the reaction, ie the page before the reaction occurred and the page after the reaction occurred .