I need help on these three chemistry problems. Please show all the work so I wil

Question

I need help on these three chemistry problems. Please show all the work so I will understand how to complete them. Thank you!

#1--- Tris(pentaflourophenyl)borane, commonly known as BARF, is a polymerization initiator used in the manufacture of plastics. It is composed entirely of C, F, and B; it is 42.23% C by mass and 55.66% F by mass.

What is the empirical formula of BARF?

#2---The vanadium in an ore sample is converted to VO2+. The VO2+ ions produced from the sample are then titrated with MnO4- in acidic solution to form V(OH)4+ and manganese(II) ion.

Write the balanced redox half-cells and final equation for the reaction occurring in the titration.

To titrate the solution, 26.45 mL of 0.02250 M MnO4- was required. How many moles of vanadium were present in the solution?

If the original sample was 58.1% V by mass, what was the mass of the ore sample?

#3---Nitromethane, CH3NO2, is burned as a fuel; the (unbalanced) equation for the reaction is

CH3NO2(l) + O2(g) → CO2(g) + N2(g) + H2O (g)

The standard enthalpy change of the reaction ∆H °rxn for the balanced reaction with the lowest whole number coefficients is -1288.5 kJ. If the ∆H °f of CO2(g) = -393.5 kJ/mol and the ∆H °f of H2O (g) = -242 kJ/mol, calculate ∆H °f for nitromethane.

A sample of nitromethane is sealed in a 15.0-L flask with oxygen gas, in the stoichiometric ratio indicated in the balanced equation above. The flask is heated to 100 °C and the reaction is then allowed to go to 100% completion. If the total pressure of all gases in the flask at 100 °C is 950 torr when the reaction is complete, what was the mass of nitromethane sample used?

Explanation / Answer

answer no 1

When you are given the % composition it is best to do the molecular weight first

moles of the BARF in solution = .01267 x 347 / 1000 = 0.00440

these weigh 2.251 g so 1 mole weighs 2.251 / 0.00440 g / mole = 511.59 g .. the molecular weight

mass of C in 1 mole = 42.23 % of 511.59 = 216.0 g or 18 moles of C

mass of F = 55.66 % of 511.59 = 284.75g or 15 moles of F

and 100 - ( 42.34 + 55.66) % of 511.59 g of B or 2 % of 511.59 = 10.23 g or 1 mole of B

molecular formula is C18F15B which is also the simplest ratio of the elements present so empirical formula is also C18F15B

answer no 2

VO2+ + 3H2O => V(OH)4+ + 2H+ + e- -1.00 V

MnO4− + 8H+ + 5e− => Mn2+ + 4H2O +1.51 V

5VO2+ + 15H2O => 5V(OH)4+ + 10H+ + 5e- X5

adding and canceling electrons

5VO2+ + 15H2O + MnO4− + 8H+ => 5V(OH)4+ + 10H+ Mn2+ + 4H2O

eliminating the common ions and water

5VO2+ + 11H2O + MnO4- => 5 V(OH)4+ + 2H+ + Mn2+

1 mole of MnO4- reacts with 5 moles of VO2+

moles of MnO4- at the eq point = 0.0225 * 26.45 / 10000 = 0.000595125

moles of VO2+ at the eq point = 5 X 0.000595125
= 0.00297 moles of VO2+ in solution
moles of V in solution = = 0.00297
0.00297 x 50.94 g of V = 0.151g

if this is 58.1 % of the sample the ore sample weighed 0.151 X 100 / 58.1 =0.260g

answer no 3

4CH3NO2(liquid) + 3O2(gas) = 4CO2(gas) + 2N2(gas) + 6H2O(gas)

Ho(reaction) = H of (products) - H of (reactants)

-1288.5 = 4 x (-393.5 ) + 0 + 6 x .242 - ( â–²H °f 4CH3NO2 + 0 )

â–²H °f (4CH3NO2) = 1288.5 -1574 + 1.45 = -284.05

â–²H °f (CH3NO2) = 284.05 / 4 = -71.0 kJ/ mole

number of moles of CH3NO2 = PV / RT
R = 62.363 L Torr K−1 mol−1

n= 950 x 15 / 62.363 x 373 = 0.613 moles or

0.613 x 61.04 = 37.42g

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