Chemistry MCQs for NEET — Practice Questions with Answers

3KMnO₄+5FeC₂O₄+24H+→5Fe³⁺+10CO2+3Mn²⁺+3K⁺+12H2O\

1. Ferrous is oxidized to ferric state
2. Carbon and hydrogen are converted to CO2 and H2O
3. All the metallic ions are converted to sulphates
4. Write the equation keeping in mind that the valency of of iron in ferric state is 3, that of K and Mn is one and two respectively as
5. FeC2O4 + KMnO4 + H2SO4 = K2SO4 + MnSO4 + Fe(SO4)3 + H2O + CO2
6. FeC2O4 is oxidized and KMnO4 is reduced. Hence the main reaction is
7. Fe2+ → Fe3+ (removal of 1 electron)

Mn7+ → Mn 2+ (addition of 5 electrons)

In order to balance multiply (Fe2+ → Fe3+) by 5

Now you know that there should be 5 FeC2O4 molecules. Write that and balance the rest.

Mole (n) = $\frac{Weight }{Molecular weight}=\frac{No. of molecules}{N_A}$

No. of molecules of  K₄[Fe(CN)₆] = 1.6 x 2 x 6 x 10²³

No. of ions = 5 x 1.6 x 2 x 6 x 10²³

= 9.6 x 10²⁴

CaCO₃(S) $\stackrel{∆}{\to }$ CO₂(g) + CaO(s)

100 kg $\to$44 kg

$\frac{80}{100}\times 10kg$ $\to$$\frac{44}{100}\times \frac{80}{100}\times 10=3.52 kg$

The reaction is: C(s) + CO2(g) → 2CO(g). Initially, there are 1 mol of CO2. After the reaction, the total volume is 1.8 L, which corresponds to 1.8 mol of gaseous product (assuming STP). Since 1 mol of CO2 produces 2 mol of CO, the moles of CO produced is 0.0714 mol.

Moles of CO₂ = $\frac{volume of gas}{22.4}=\frac{2.8}{22.4}$=0.125 mol

moles of oxygen atom in given CO₂ = 0.125 x 2=0.25 mol

1 mol N^-3 ion $\to$1 x 10 mol e^- 

1 mol O^-2 ion $\to$1 x 10 mol e^- = 6.023x10^-24e^-

1 mol CH₄ $\to$1 x 10 mol proton

1 mol H₂O $\to$1 x 10 mol proton

Balanced Equation:

24C + 22H₂ + 11O₂ -> 2C₁₂H₂₂O₁₁

moles for reaction: 

7(84/12), 6(12/2), (56/22.4)

dividing by coefficients in balanced equation & comparing: 

7/24, 6/22,  (56/22.4)/11 

0.29, 0.27, 0.22

Oxygen is the limiting reagent. So, weight of sucrose obtained = 0.22 * 2(coefficient for sucrose in reaction) * 342(sucrose molecular weight) = 155.5g

Only NaHCO₃ evolve CO₂ gas. Na₂CO₃ does not evolve CO₂ gas even on red heating.

So, the only reaction happening is: 

2NaHCO₃ -> Na₂CO₃ + CO₂ + H₂O

Weight: 

2(23+1+12+16*3) -> (2*23+12+16*3) + (12+16*2) +  (1*2+16)
                               
    168g                 ->         102g          +       44g      +    18g

Also, CO₂ mass as per 56ml volume obtained at STP = 56/(22.4 * 1000) * (44) = 0.11g

So, NaHCO₃ weight = 0.11/44*168 = 0.42g (using ratio of weights from equation)

Therefore, original Na₂CO₃ weight in the original mixture = 10 - 0.42 = 9.58g

Equivalents of KMnO₄ = equivalent of FeSO₄ + equivalent of FeC₂O₄

x × 5=1 × 1 + 1 × 3

x = $\frac{4}{5}$ mole