-Help is on pg.2 and answers are at the end to check overQuestions 1,2 and 3 pertain to the information given below.
X is dilute Hydrochloric acid containing 3.65g/dm3 of HCl of the solution.
Y is a Sodium Hydroxide solution containing 5.0g/dm3 of solution.
Question1
1. How many moles of HCl does X contain?
2. What is the concentration of X given as mol/dm3?
Question2
The reaction of X with Magnesium Hydroxide is given by the equation below:
2HCl
+
Mg(OH)2

MgCl2 +
2H2O
Calculate:
1. The number of moles of MgCl2 yielded when 1 dm3 of X reacts with excess Mg(OH)2?
2. The mass of MgCl2 yielded when 1 dm3 of X reacts with Mg(OH)2?
3. The number of moles of MgCl2 yielded when 100 cm3 reacts with excess Mg(OH)2?
Question 3
1.
2.
3.
4.
Write the balanced equation of the reaction of Y with Sulphuric acid.
Calculate the number of moles of H2SO4 which will react with 1 dm3 of Y.
Calculate the number of moles of the salt produced by this neutralization reaction.
Calculate the mass of the salt produced by this neutralization reaction.
Guide to questions 1&2:
Question1:
1. According to the question the solution contains 3.65g of HCl with every dm3 so if were change
the 3.65g mass to moles you’ll get 0.1 moles of HCl.
2. The 3.65g was already given as part of the 1 dm3 solution so when you find the number of
moles present it will already be as part of a 1dm3 solution so the concentration will be
0.1mol/dm3.
Question2:
1.
X contains Hydrochloric acid or HCl, the question has given the number of moles present with
each dm3 as 0.1mol/dm3 so there will be 0.1 moles of HCl present with the 1dm3 of solution.
According to the equation the ratio of HCl to MgCl2 produced is 2 to 1 so with each mole of
MgCl2 being produced there would be twice that number of moles of HCl to start with, or it
might also be said that with each mole of HCl reacting there would be half as much MgCl2
product, they both mean the same thing.
So according to the mole ratio of HCl to MgCl2 shown by the equation 0.1 moles of HCl will
yield half the number of moles of MgCL2, so from the 0.1 moles of HCl 0.05 moles of MgCl2 is
produced.
2. From question 1 0.05 moles of MgCl2 are yielded so using moles equation number 1, the 0.05
moles of MgCl2 is equal to :
0.05 moles x 95g(mass of one mole of MgCl2) = 4.25g
3.
The first thing to calculate is the number of moles present with 100cm3 of the Hydrochloric
acid. Using the molar concentration formula (1000cm3/ given volume) x number of moles it is
found that there will be 0.01 moles of HCl present. Using the mole ratio shown by the equation
the 0.01 moles of HCl yields half as much MgCl2 or 0.005 moles of MgCl2.
Do Not Scroll below
Until you’ve
completed questions
1-3. Use the help
provided after
question 3 to help
you with the
problems.
Answers:
Question1:
1.
0.1 moles
2.
0.1 mole/dm3
Question2:
1.
0.05 moles
2.
4.25g
3.
0.005 moles
Question 3:
1.
H2SO4 +
2NaOH
2.
0.625 moles of H2SO4
3.
0.625 moles of Na2SO4
4.
8.875g of Na2SO4
Na2SO4
+
2H2O