IB DP Chemistry Topic 1.3 Reacting masses and volumes SL Paper 1

Question

What is the maximum mass, in g, of magnesium oxide that can be obtained from the reaction of oxygen with 2.4 g of magnesium?

A.     2.4

B.     3.0

C.     4.0

D.     5.6

▶️Answer/Explanation

C

\[{\text{2Mg(s)}} + {{\text{O}}_2}({\text{g)}} \to {\text{2Mg}}{{\text{O}}}({\text{g)}}\]

Given, 2.4 g of Mg

Number of moles of Mg = (Mass of Mg) ÷ (Molar mass of Mg)

i.e. Number of moles of Mg = 2.4/24 = 0.1 moles of Mg

From the equation above, 2 moles of Mg gives 2 moles of MgO. 

Hence, 0.1 mole of Mg will give 0.1 mole of MgO. 

Maximum mass of MgO that can be obtained = 0.1×(Molecular mass of MgO) = 0.1×(24+16) = 4.0 g

Question

\({\text{5 d}}{{\text{m}}^{\text{3}}}\) of carbon monoxide, CO(g), and \({\text{2 d}}{{\text{m}}^{\text{3}}}\) of oxygen, \({{\text{O}}_{\text{2}}}{\text{(g)}}\), at the same temperature and pressure are mixed together. Assuming complete reaction according to the equation given, what is the maximum volume of carbon dioxide, \({\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\), in \({\text{d}}{{\text{m}}^{\text{3}}}\), that can be formed?

\[{\text{2CO(g)}} + {{\text{O}}_2}({\text{g)}} \to {\text{2C}}{{\text{O}}_2}({\text{g)}}\]

A.     3

B.     4

C.     5

D.     7

▶️Answer/Explanation

B

According to Gay Lussac’s law:

2CO(g) + O2(g)  →  2CO2(g) 

2 vol         1 vol            2 vol

1 vol         0.5 vol         1 vol

1 vol of CO requires 0.5 vol of oxygen.

5 dm3 CO requires 2.5 dm3 of oxygen. Here, we have only 2 dmof oxygen. Here, oxygen is the limiting reagent.

0.5 vol of oxygen will form 1 vol of CO2

So, 2 dmof oxygen will form 4 dmof CO2.

Question

The volume of an ideal gas at 27.0 °C is increased from \({\text{3.00 d}}{{\text{m}}^{\text{3}}}\) to \({\text{6.00 d}}{{\text{m}}^{\text{3}}}\). At what temperature, in °C, will the gas have the original pressure?

A.     13.5

B.     54.0

C.     327

D.     600

▶️Answer/Explanation

C

Using ideal gas equation, we have

\(\frac{P1V1}{T1} = \frac{P2V2}{T2}\)  where P=Pressure, V=Volume and T=Absolute temperature of the gas.

Given, P1=P2=P(say) (Since pressure is constant)

V1=V(say), where V is the initial volume of the gas.

V1 = 3 dm3 , V2 = 6 dm3

Therefore, V2=2V since final volume is double the initial volume.

T1=(27+273)K=300K

T2=?

Substituting the values obtained in the gas equation, we have

\(\frac{PV}{300} = \frac{P(2V)}{T2}\)

Solving, we get T2=final temperature of the gas=600K or (600–273)degree C or 327°C.

Question

5.0mol of Fe2O3(s) and 6.0mol of CO(g) react according to the equation below. What is the limiting reactant and how many moles of the excess reactant remain unreacted?

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

▶️Answer/Explanation

B

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

1               :    3

5 mole Fe2O3 requires 15 moles of CO.

We have only 6 moles of CO. Hence, CO is the limiting reactant. Fe2O3 is the excess reactant. 

3 moles of CO require 1 mole of Fe2O3

So, 6 moles of CO will require 2 mole of Fe2O3.

Hence, moles of excess reactant remaining = 5 – 2 =3.

Question

Which electron transition emits energy of the longest wavelength?

▶️Answer/Explanation

A

Wavelength is given by : 

\(\frac{1}{\lambda}=RZ^2(\frac{1}{n_2^2} – \frac{1}{n_1^2})\) 

Where λ = wavelength, R = Rydberg constant, Z = atomic number and n1 and n2 are principle quantum number.

For A, n= 3, n1 = 5

\(\frac{1}{\lambda_1}=RZ^2(\frac{1}{3^2} – \frac{1}{5^2})\) 

\(\frac{1}{\lambda_1}=RZ^2(\frac{1}{9} – \frac{1}{25})\) 

\(\lambda _1 = \frac{14.0625}{RZ^2}\)

For B, n= 5, n1 = 4

In this case, energy emission is not possible since emission occurs only from higher n to lower n. 

For C, n= 3, n1 = 2

In this case, energy emission is also not possible since emission occurs only from higher n to lower n. 

For D, n= 1, n1 = 3

\(\frac{1}{\lambda_4}=RZ^2(\frac{1}{1^2} – \frac{1}{3^2})\) 

\(\lambda _4 = \frac{1.125}{RZ^2}\)

Hence, energy of longest wavelength is emitted in case of electron A. 

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