6. How many milliliters of 0.238 M KMno, are needed to react with 3.36 g of iron

Question

6. How many milliliters of 0.238 M KMno, are needed to react with 3.36 g of iron (I1) sulfate, Feso,7 The reaction is as follows 10Feso, (aq) +2KMno,(aq)+8H,SO,(aq)>5Fe,(SO,),(aq) 2MnSO, (aq) K,So (ag) +8H,0(1) 7. A 1.28-g sample of a colorless liquid vas vaporized in a 250-mL flask at 121 C and 786 mmHg. What is the molecular veight of this substance? 8. Sma11 amounts of hydrogen are conveniently prepared by reacting zine with hydrochloric acid. Zn (s) + 2HC1 (aq)->ZnCl,(aq) +H(g How many grams of zinc are required to prepare 2.50 L H, gas at 765 mmHg and 22 C?

Explanation / Answer

Soluion:- (6) The given balanced equation is..

10FeSO4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ----> 5Fe2(SO4)3(aq) + 2MnSO4(aq) + K2SO4(aq) + 8H2O(l)

It is a stoichiometry problem, we would calculate the moles of Iron(II)sulfate from it's given grams and then using mol ratio from the balanced equation we would calculate the moles of KMnO4 and finally it's volume could be calculated using it's molarity and moles.

3.36 g FeSO4 x (1 mol FeSO4/151.9 g FeSO4) x (2 mol KMnO4/10 mol FeSO4) x (1L KMnO4/0.238 mol KMnO4)

= 0.0186 L KMnO4

These liters could be converted to mL.

0.0186 L x (1000 mL/1L) = 18.6 mL

(7) It's based on ideal gas law equation..

PV = nRT

in this equation n stands for moles which is mass/molar mass. if we represent mass by w and molar mass by M then..

n = w/M

So, PV = wRT/M

we are asked to calculate the molecular weght that is M. So...

M = wRT/PV

given, w = 1.28 g, V = 250 mL = 0.25 L

T = 121 + 273 = 394 K

P = 786 mmHg x (1 atm/760 mmHg) = 1.03 atm

let's plug in the values..

M = (1.28 g x 0.0821 atm.L.mol-1.K-1 x 394 K)/(1.03 atm x 0.25 L)

M = 160.79 g.mol-1

(8) The given balanced equation is...

Zn(s) + 2HCl(aq) -----> ZnCl2(aq) + H2(g)

using ideal gas law equation we can calculate the moles of H2, using mol ratio the moles of Zn and then it's grams could be calculated.

V = 2.50 L

P = 765 mmHg x (1 atm/760 mHg) = 1.01 atm

T = 22 + 273 = 295 K

n = PV/RT

n = (1.01 atm x 2.50 L)/(0.0821 atm.L.mol-1.K-1 x 295 K)

n = 0.104 mol H2

0.104 mol H2 x (1mol Zn/1 mol H2) x (65.38 g Zn/1mol Zn) = 6.80 g Zn

(9) Partial pressure of oxygen gas = 0.200 atm

total pressure of the gas mixture = 3.00 atm

according to Dalton's law of partial pressure. "Total pressure of the gas mixture is the sum of partial pressures of the gases present in it."

So, Total pressure = partial pressure of oxygen + partial pressure of helium

partial pressure of helium = 3.00 atm - 0.200 atm = 2.80 atm

Volume = 1.00 L

T = 20 + 273 = 293 K

molar mass of Helium, M = 4.0026 g.mol-1

PV = wRT/M

PVM = wRT

w = PVM/RT

w = (2.80 atm x 1.00 L x 4.0026 g.mol-1)/(0.0821 atm. L. mol-1.K-1 x 293 K)

w = 0.466 g Helium

(10) 0 degree C of ice is changing to 100 degree steam. It's a multistep problem and would be solved as..

1st step:- 0 degree C ice is converted to 0 degree water

q1 = m*delta Hfus

m = 20.0 g

delta Hfus = 334 J/g

so, q1 = 20.0 g x 334 J/g = 6680 J

2nd step:- 0 degree C ice is changed to 100 degree C water.

q2 = m c delta T

m = 20.0 g

c = 4.18 J/g.C

delta T = 100 - 0 = 100 C

q2 = 20.0 g x 4.18 J/g.C x 100 C = 8360 J

3rd step:- 100 degree C water is changed to 100 degree C steam.

q3 = m* delta Hvap

q3 = 20.0 g x 2.25 kJ/g x 1000J/1kJ = 45000 J

total required heat = q1 + q2 + q3

total required heat = 6680 J + 8360 J + 45000 J = 60040 J OR 60.0 kJ

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