Question

Questions 15-17 utilize the following balanced chemical equation: 2 K2CrOdaq) + 3 SO2(g) + 4 HCl(aq) → Cr2(SO4)3(s) + 4 KCl(aq) + 2 H2011) 15. Toxic chromate can be precipitated from an aqueous solution by bubbling SO2(g) through the solution How many moles of the gas are required to treat 3.00 x 10L of a solution that is 3.70 x 10 Min potassium chromate (KaCrO4) according to the above reaction? A. 2250 mol B. 1000 mol C. 1670 mol D. 3750 mol E. 750 mol in 16. Determine the theoretical yield (in grams) of chromium(VI) sulfate (Cr2(SO4)s) when 50.0 mL of 0.625 M of K CrO4 reacts with when 2.75 g of sulfur dioxide according to the above reaction Cr:(SO.). = 392.16 g/mol; SO2 = 64.07 g/mol) Md No 1 Er / cm 1100 Fm 1291/6791/ 2 Ho 67 / Es A. 5.618 B. 5.10 g C. 1.23 g D. 0.754g E. 6.13 g 103 C 66 Dy 96 Tb 65 cl BK2511 EU15191 $1096 64 Gd 276 63 Eu 270 Sm 62 17. Determine the number of moles of excess reactant that remains after 0.600 mol of K Croa reacts with when 0.500 mol of sulfur dioxide according to the above reaction. on 272 61 Sg 271 A. 0.333 mol B. 0.100 mol C. 0.150 mol D. 0.400 mol E. 0.267 mol Db 268 Rf 076€ Rb 1 mol BCI 172 37

Answer 1

Answer:-

(15)-

Given:-

volume of K₂CrO₄ solution = 3.00 $\times$ 10⁸ L

molarity of K₂CrO₄ solution = 3.70 $\times$ 10^-6 M = 3.70 $\times$ 10^-6 mol/L

No. of moles of SO₂ gas (n_SO2 ) = ?

As we know that

molarity of compound (M) = no. of moles of compound (n) / volume of solution (L)

no. of moles of compound (n) = molarity of compound (M) $\times$ volume of solution (L)

therefore

No. of moles of K₂CrO₄ (n_K2CrO4 ) = molarity of K₂CrO₄ solution $\times$ volume of K₂CrO₄ solution

No. of moles of K₂CrO₄ (n_K2CrO4 ) = 3.70 $\times$ 10^-6 mol/L $\times$ 3.00 $\times$ 10⁸ L

No. of moles of K₂CrO₄ (n_K2CrO4 ) = 11.1 $\times$ 10² mol = 1110 mol

Also we know that

2K₂CrO_4(aq) + 3SO_2(g) + 4HCl_(aq)  $\rightarrow$ Cr₂(SO₄)_3(s) + 4KCl_(aq) + 2H₂O_(l)

2 mol 3 mol 4 mol 1 mol 4 mol 2 mol

As mentioned in above reaction

2 moles of K₂CrO₄ reacts with = 3 moles of SO₂ gas

1 mol of K₂CrO₄ reacts with = 3 / 2 moles of SO₂ gas

then

1110 moles of K₂CrO₄ reacts with = 3 $\times$ 1110 / 2 moles of SO₂ gas

1110 moles of K₂CrO₄ reacts with = 3330 / 2 moles of SO₂ gas

1110 moles of K₂CrO₄ reacts with = 1665 moles of SO₂ gas

therefore

No. of moles of SO₂ gas required (n_SO2 ) = 1665 moles of SO₂ gas (i.e the answer)

So correct option is 'C' i.e 1670 moles which is approximately equal to the 1665 moles of SO₂ gas.

(16)-

Given:-

molar mass of SO₂ gas = 64.07 g /mol

mass of of SO₂ gas = 2.75 g

molar mass of Cr₂(SO₄)₃ = 392.16 g /mol

volume of K₂CrO₄ solution = 50.0 mL = 50.0 / 1000 = 0.05 L

molarity of K₂CrO₄ solution = 0.625 M = 0.625 mol /L

As we know that

No. of moles of K₂CrO₄ (n_K2CrO4 ) = 0.625 mol /L $\times$ 0.05 L

No. of moles of K₂CrO₄ (n_K2CrO4 ) = 0.03125‬ mol

Also we know that

No. of moles of compound (n ) = mass of compound (g) / molar mass of compound (g/mol)

then

No. of moles of SO₂ gas (n_SO2 ) =mass of SO₂ gas (g) / molar mass of SO₂ gas (g/mol)

No. of moles of SO₂ gas (n_SO2 ) = 2.75 g / 64.07 g /mol

No. of moles of SO₂ gas required (n_SO2 ) = 0.04292 mol

Since

2K₂CrO_4(aq) + 3SO_2(g) + 4HCl_(aq)  $\rightarrow$ Cr₂(SO₄)_3(s) + 4KCl_(aq) + 2H₂O_(l)

2 mol 3 mol 4 mol 1 mol 4 mol 2 mol

2 $\times$ 194.19 g 3 $\times$ 64.07 g 392.16 g /mol

388.38 g 192.21 g   392.16 g /mol

As we know that

3 moles of SO₂ gas reacts with = 2 moles of K₂CrO₄

1 mol of SO₂ gas reacts with = 2 / 3 moles of K₂CrO₄

then

0.04292 moles SO₂ gas reacts with = 2 $\times$ 0.04292 / 3 moles of K₂CrO₄

0.04292 moles SO₂ gas reacts with = 0.08584‬ / 3 moles of K₂CrO₄

0.04292 moles SO₂ gas reacts with = 0.02861 moles of K₂CrO₄

So

unreacted no. of moles of K₂CrO₄ = 0.03125‬ moles - 0.02861 moles

unreacted no. of moles of K₂CrO₄ = 0.00264‬ moles

As mentioned in above reaction No. of moles of SO₂ gas (n_SO2 ) is acts as limiting reagent or agent which determine the overall yield of the  Cr₂(SO₄)₃ .

therefore

3 moles of SO₂ gas produced = 392.16 g of Cr₂(SO₄)₃

1 mol of SO₂ gas produced = 392.16 / 3 g of Cr₂(SO₄)₃

then

0.04292 moles of SO₂ gas produced = 0.04292 $\times$ 392.16 / 3 g of Cr₂(SO₄)₃

0.04292 moles of SO₂ gas produced = 16.8315072‬ / 3 g of Cr₂(SO₄)₃

0.04292 moles of SO₂ gas produced = 5.61 g of Cr₂(SO₄)₃

So theoretical yield of the  Cr₂(SO₄)₃ = 5.61 g of Cr₂(SO₄)₃ (i.e the answer)

The correct option is 'A' i.e 5.61 g.

​​​​​​​(17)-

Given:-

given no. of moles of SO₂ gas (n_SO2 ) = 0.500 mol

given no. of moles of K₂CrO₄ (n_K2CrO4 ) = 0.600 mol

As we know that

2K₂CrO_4(aq) + 3SO_2(g) + 4HCl_(aq)  $\rightarrow$ Cr₂(SO₄)_3(s) + 4KCl_(aq) + 2H₂O_(l)

2 mol 3 mol 4 mol 1 mol 4 mol 2 mol

Also we know that

3 moles of SO₂ gas reacts with = 2 moles of K₂CrO₄

1 mol of SO₂ gas reacts with = 2 / 3 moles of K₂CrO₄

then

0.500 moles SO₂ gas reacts with = 2 $\times$ 0.500 / 3 moles of K₂CrO₄

0.500 moles SO₂ gas reacts with = 1.0 ‬ / 3 moles of K₂CrO₄

0.500 moles SO₂ gas reacts with = 0.333 moles of K₂CrO₄

therefore

No.of moles of K₂CrO₄ that reacted (n'_K2CrO4 ) = 0.333 moles

unreacted excess no. of moles of K₂CrO₄ = given no. of moles of K₂CrO₄ (n_K2CrO4 ) - No.of moles of K₂CrO₄ that reacted (n'_K2CrO4 )

unreacted excess no. of moles of K₂CrO₄ = 0.600 moles - 0.333 moles

unreacted excess​​​​​​​ no. of moles of K₂CrO₄ = 0.267 mol (i.e the answer)

The correct option is 'E' i.e 0.267 mol of K₂CrO₄ .