Moles Calculations

Tullia SG

If someone could be good enough to answer the following questions with full method shown, then that would be appreciated. I've just completed empirical and molecular formulas and also the first three questions of this worksheet, but I've only jotted down some answers for the following and I'm unsure. I'm wanting to understand and check my working and answers more than anything, so I can tackle questions like this comfortably without it taking me hours.

Thanks in advance.

4) The pollutant sulphur dioxide can be removed from the air by the following reaction. What mass of sulphur dioxide will 10.0 kg of calcium carbonate remove?

2 CaCO₃ + 2 SO₂ + O₂ ® 2 CaSO₄ + 2 CO₂

5) What mass of Na₂O is produced when 2.50 g of sodium is burned in oxygen?

4 Na + O₂ ® 2 Na₂O

6) What mass of magnesium oxide is formed when 6.00 g of magnesium reacts with oxygen?

2 Mg + O₂ ® 2 MgO

7) What mass of carbon dioxide is produced when 5.60 g of butene (C₄H₈) is burnt?

C₄H₈ + 6 O₂ ® 4 CO₂ + 4H₂O

8) What mass of oxygen is required to oxidise 10.0 g of ammonia to NO?

4 NH₃ + 5 O₂ ® 4 NO + 6 H₂O

9) In the Solvay process, ammonia is recovered by the reaction shown below. What is the maximum mass of ammonia that can be recovered from 2.00 tonnes of ammonium chloride and 0.500 tonnes of calcium oxide (quicklime)?

2 NH₄Cl + CaO ® CaCl₂ + H₂O +2NH₃

10) When 6.20 g of Na₂CO₃.nH₂O is heated gently until no further change in mass occurs, to remove the water of crystallisation, 5.30 g of anhydrous sodium carbonate remained. Work out the formula mass of the Na₂CO₃.nH₂O, and so the value of n.

11) A mixture of MgSO₄.7H₂O and CuSO₄.5H₂O is heated at 120°C until a mixture of the anhydrous compounds is produced. If 5.00 g of the mixture gave 3.00 g of the anhydrous compounds, calculate the percentage by mass of MgSO₄.7H₂O in the mixture.

Darkness_hacker99 SG

Dear Tullia,

Welcome to Homework Forum. I am the moderator here and will assist you in your learning today. I will first share with you my suggested solutions then share with you some of the important note on Mole concepts.

Step 1: Write the chemical equation

Q4)

No. of mol of CaCO3 = 10000/100 = 100mol

Ratio of CaCO3 : SO2 = 2 : 2 = 1: 1

This means 100 mol of CaCO3 will react with 100 mol of SO2.

Mass of SO2 removed is 100 mol x Ar SO2 = 100 x 64 = 6400g or 6.4kg.

Q5)

No. of mol of Na = 2.50g/23 = 0.108695.... , we'll take 0.109 mol.

Ratio of Na : Na2O = 4 : 2 = 2 : 1

This means 4 mol of Na burned will produce 2 mol of Na2O. (or it means 2 parts of Na will produce 1 part of Na2O)

Mass of Na2O produced = [0.109 x 2/4] x 62 = 3.379 g

Q6)

No. of mol of Mg = 6/24 = 0.25 mol

Ratio of Mg : MgO = 2 : 2 = 1:1

This means 2 mol of Mg will react to produce 2 mol of MgO (or it means 1 part of Mg to 1 part of MgO)

Mass of MgO produce = 0.25 x Ar MgO = 15g

Q7)

No. of mol of C4H8 = 5.6/56 = 0.1 mol

Ratio of C4H8 : CO2 = 1 : 4

This mean 1 mol of C4H8 will react to produce 4 mol of CO2.

No. of mol of CO2 produced will be 0.4 mol

Mass of CO2 = 0.4 x Ar CO2 = 17.6g

Q8)

No. of mol of NH3 = 10/ 17 = 0.5882.. we'll take 0.588 mol

Ratio of NH3 : O2 = 4:5

This means 4 mols of NH3 will react with 5 mols of O2.

Mass of O2 = [0.588 x 5/4] x 32 = 23.52g

Q9 - Limiting reactant

Step1 : convert all values to mol

No. of mol of NH4Cl = 2,000,000 / 53.5 = 37383.1775... we will take 37383 mol

No. of mol of CaO = 500,000 / 56 = 8928.5714... we will take 8929 mol

Step2 : compare ratio and identify limiting reactant

Ratio of NH4Cl : CaO = 2:1

That means 2 mols of NH4Cl will react with 1 mol of Cao.

8929 mol of CaO would require 8929 x 2=17858 mol of NH4Cl to react with. However there is 37383 mol of NH4Cl, which is greater than the required 17858 mol of NH4Cl.

From here, we can tell that NH4Cl is in excess. CaO is the limiting reactant which will be used up first in the reaction.

Step3: Calculation

When doing Limiting Reactant questions, always use the limiting reactant as the means to calculate the mass of end product.

Ratio of CaO : NH3 = 1:2

Mass of NH3 recovered = (8929 x 2/1) x 17 = 303586g or 303.586 kg,

Darkness_hacker99 SG

Q10 - Water of Crystalization

Mass of anhydrous salt = 5.30g, mass of water driven = 6.2g - 5.3g = 0.9g

% of water of crystalization in the crystal = 0.9/6.2 x 100 = 14.52 %

Mass ratio of Na2CO3 : H2O = 5.30 : 0.90

[Ar Na=23, C=12, O=16, H=1]

Convert Mass ratio to Mol ratio = 5.30/106 : 0.90/18

which is 0.05 : 0.05 or 1:1, therefore formula of hydrated salt is Na2CO3. H2O

Chemfreak022 SG

For question 11,

mass of water evolved = 2.00 g

Can use algebra to solve the problem

Let x g be mass of MgSO4.7H2O and (5-x) g be mass of CuSO4.5H2O

Then

(7x18)x/246.97 + (5x18)(5-x)/249.7 = 2 Equation for water loss

Solve for x

x = 1.31 g (after solving for x)

Should be able to get the final ans from here.

Tullia SG

Thank you very much, Darkness. And to you also, Chem.

Darkness, I'd done reasonably well when checking against your working out and answers, and you've definitely helped improve my approach to such questions. Getting a balanced equation to begin with and then finding the moles and subsequent ratio and working from there as needed has improved my working out and can only assist me in the future.

Question 9, 10 & 11 are where I made some slight errors, but I now see where I've gone wrong with them and have redone the questions and gone through each step so I understand it. It's the wording that gets me, as I've said before.

Again, thank you very much.

Darkness_hacker99 SG

In O level Chemistry, you are to construct the balanced chemical equation yourself.

For limiting reactant, please refer to past threads:

http://sgforums.com/forums/2258/topics/177358 (this thread I created when I was studying limiting reactant) , refer to yuko-ogura's explanation.