Chems101 experiment 1

Chemy 101 Experiment..1 Data Sheet The Equation for A Reaction
NAME : Musherah Moqbel Ali Alzoubah ID: 202002276 SEC:–18—-AIM :
1) To determine the molar ratio between sodium hydrogen carbonate
and hydrochloric acid when they react.
2) To write an equation for the reaction.
A:Preliminary Test :
NaHCO3 (s) + HCl (aq)
NaCl(aq) + CO2(g) + H2O(l)
Observations:
1-a What happened to lime water ? It turns milky.
Why ? CO2 react with lim water and forms white precipitate of calcium carbonate
(CaCO3(s)) .
1-b Write a complete equation for reaction
CO2(g)
+
Ca (OH )2 (aq)
CaCO3(s)
+
H2O(l)
2- What happened to the lit splint ? The lighted splint turned off.
Why ? Because CO2(g) is not support combustion (burning).
B: Equation of a reaction
1-Mass of empty evaporating dish
m1 = 33.1421 g
2. Mass of evaporating dish with +NaHCO3 m2 =33.7353 g
Mass of NaHCO3 (m2-m1) = 33.7353 – 33.1421 = 0.5932
3- Mass of evaporating dish and (NaCl) after heating(m3) = 33.5833 g
4-Mass of residue(NaCl) (m3-m1) = 33.5833 – 33.1421 = 0.4412 g
5-Moles Of NaHCO3= MassNaHCO3 / M.MNaHCO3=
M.MNaHCO3= (22.990) +( 1.008) + (12.011 )+ (3×15.999) = 84.006 ≈ 84.01
MassNaHCO3 / M.MNaHCO3=
𝟎.𝟓𝟗𝟑𝟐
= 7.061× 𝟏𝟎−𝟑 mole
𝟖𝟒.𝟎𝟏
6-Moles of NaCl = Mass NaCl / MMNaCl MMNaCl = 22.990 + 35.453 = 58.443
𝟎.𝟒𝟒𝟏𝟐
𝟓𝟖.𝟒𝟒𝟑
= 7.549× 𝟏𝟎−𝟑 mole
7- Moles of NaHCO3 :
𝟕.𝟎𝟔𝟏×𝟏𝟎−𝟑
Moles of NaCl
𝟕.𝟓𝟒𝟗×𝟏𝟎−𝟑
:
𝟕.𝟎𝟔𝟏×𝟏𝟎−𝟑
𝟕.𝟎𝟔𝟏×𝟏𝟎−𝟑
1
:
1.06≈1
1
:
1
Moles of NaHCO3 :
1
Moles of HCl
:
1
8- The balanced equation is
Na𝐂𝐥(𝐚𝐪) + C𝐎𝟐 (𝐠) + 𝐇𝟐 𝐎(𝐥)
NaHCO3(s) + HCl (aq)
9- Test of Cl
10 .Observation-The colour of precipitate is white
NaCl +
Ag𝐂𝐥(𝐬) + 𝐍𝐚𝐍𝐨𝟑(𝐚𝐪)
AgNO3
1.Calculate the number of moles in the following
1)0.4544g Na2CO3:
M.M Na2CO3 =(2 x 22.990) + (12.011) + (3 x 15.999) = 105.988≈ 105.99
𝑀𝑎𝑠𝑠𝑁𝑎2 𝐶𝑜3 / 𝑀𝑀𝑁𝑎2 𝐶𝑜3 =
2)0.2075 g NaCl
𝟎.𝟒𝟓𝟒𝟒
𝟏𝟎𝟓.𝟗𝟗
= 4.287× 𝟏𝟎−𝟑 mole
M.M NaCl =(22.990) + (35.453) = 58.443
𝑀𝑎𝑠𝑠𝑁𝑎𝐶𝑙 / 𝑀𝑀𝑁𝑎𝐶𝑙 =
𝟎.𝟐𝟎𝟕𝟓
𝟓𝟖.𝟒𝟒𝟑
= 3.550× 𝟏𝟎−𝟑 mole
CONCLUSION—–We do the experiment to find and to write a chemical equation for a particular
reaction and to find the molar ratio between the reactants.
The reaction are sodium hydrogen carbonate (NaHCO3) , hydrochloric acid (HCl).
The experiment has two parts , part one is the preliminary test or gas test, part two
is for the equation for the reaction.
NaHCO3(s) + HCl (aq)
Na𝐂𝐥(𝐚𝐪) + C𝐎𝟐 (𝐠) + 𝐇𝟐 𝐎(𝐥)
We find the product, which is carbon dioxide (CO2) by using lim water, and sodium
chloride (NaCl) by using silver nitrate, and water (H2O)
Solution by finding the mass of sodium chloride , we find the molar rstio between
sodium hydrogen carbonate and (HCl).
…