Element M is prepared industrially by a two-step procedure according to the foll

Question

Element M is prepared industrially by a two-step procedure according to the following (unbalanced) equation: M_2O_3(s) + C(s) + Cl_2(g) - rightarrow MCl_3(l) + CO(g) MCl_3 (l) + H_2(g) - rightarrow M(s) + HCl(g) Assume that 0.855 g of M_2O_3 is submitted to the reaction sequence. When the HCl produced in step(2) is dissolved in water and titrated with 0.511 M NaOH, 144.2 mL of the NaOH solution is required to neutralize the HCl. (a) Balance both equations, (b) What is the atomic mass of element M, and what is its identity? (c) What mass of M in grams is produced in the reactions? When a student mixes 50 mL of 1.0 M HCI and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 degree C to 27.5 degree C. Calculate the enthalpy change (Delta H) for the reaction in kJ/mol HCl. (Assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g middot K.)

Explanation / Answer

The reactions are M2O3 +3C+3Cl2-----à2MCl3(l) + 3CO(g) Balanced 1st reaction-1

2MCl3 + 3H2------à2M +6HCl balanced second reaction-2

The reaction between HCl and NaOH is HCl+ NaOH--àNaCl+ H2O

Moles of HCl= moles of NaOH= molarity* Volume(L)= 0.511*144.2/1000 =0.0736

From the reaction, 6 moles of HCl and 2 moles of M requires 2 mole of MCl3 ( from reaction-2)

Moles of M= 0.0736 ( same as MCl3)

0.0736 moles of HCl require 0.0736*2/6= 0.02456MCl3

2 mole of MCl3 requires   1 mole of M2O3 ( from reaction-1)

0.02456 moles of MCl3 require 0.02456/2=0.01228 moles

Moles= mass/molar mass, molar mass of M2O3 = mass/moles =0.855/0.01228=69.6

Atomic mass of M*2+3*atomic mass of O= 2*M+48= 69.6, M=10.8

The metal is Boron.

Mass of M = moles* molar mass = 0.0736*10.8= 0.795 gm of M

2. total volume of solution = 50+50 =100 ml, density =1 g/ml, mass of solution =100*1= 100 gm

enthalpy change= -mass * specific heat * temperature difference= -100*4.184*(27.5-21)=- 2720 joules= -2.720 Kj

enthalpy change -ve suggests heat is evolved during the reaction.

moles of HCl in 50ml of 1M= molarity* Volume in liters= 1*50/1000 =0.05 moles

enthalpy change / mole = -2.72/0.05= -54.4 Kj/mole.

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