Assume the complete reaction of 8.22 g of O 2 in the following reaction to deter

Question

Assume the complete reaction of 8.22 g of O2 in the following reaction to determine the other quantities.

7 O2 + 4 Mn ? 2 Mn2O7

(a) moles of O2 reacting
__________mol

(b) moles of Mn required
__________mol

(c) moles of Mn2O7 formed
__________mol

(d) mass of Mn2O7 formed
__________g

________________________________________

How many moles of CF4 can be produced when 5.31 mol C reacts with 5.44 mol F2?
__________mol

How many moles of each reactant are left over? (Note that at least one of the reactants must disappear.)

Explanation / Answer

moles O2 = 8.2 g / 32.00 g/mol
= 0.25625 moles
= 0.2562 moles



The balanced equation shows
7 moles O2 requaires 4 moles Mn
So 1 mole O2 needs 4/7 moles Mn
Thus 0.25625 moles O2 will need (4/7 x 0.25625) moles Mn
= 0.146428 moles Mn needed
= 0.1464 moles Mn (4 sig figs)



7 moles O2 reacts to produce 2 moles Mn2O7
So 1 mole O2 will form 2/7 moles Mn2O7
Thus 0.871875 moles O2 will form (2/7 x 0.25625) moles Mn2O7
= 0.073214 moles Mn2O7
= 0.0732 moles (4 sig figs)


mass = molar mass x moles

mass Mn2O7 = 221.87 g/mol x 0.073214 moles
= 16.24399 g

b.

C + 2F2 --> CF4

5.31 mol C * (1 mol CF4 / 1 mol C) = 5.31 mol CF4
5.44 mol F2 * (1 mol CF4 / 2 mol F2) = 2.72 mol CF4

2.72 moles is less, so F2 is the limiting reactant.
2.72 mol CF4 is produced.

after the reaction,

left, C 5.31 - 2.72 = 2.59 mol

F2 is 0 mol

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