Dimethylnitrosamine (CH3)2N2O, is a carcinogenic formed in foods from nitrite ions used as preservatives. Calculate each of the following for this compound. Show all work for full credit. a. molar mass b. What is the % N

A. 74g/mol

B. 37.8%

Explanation:

A. Determination of the molar mass of dimethylnitrosamine, (CH3)2N2O.

The molar mass of dimethylnitrosamine, (CH3)2N2O can be obtained by adding the atomic masses of the elements present in the compound. This is illustrated below:

Molar mass of (CH3)2N2O = 2[12+ (3x1)] + (2x14) + 16

= 2[12 + 3] + 28 + 16

= 2[15] + 28 + 16

= 30 + 28 + 16

= 74g/mol.

B. Determination of the percentage composition of N in the compound. This is illustrated below:

There are 2 atoms of N in (CH3)2N2O. Mass of N in (CH3)2N2O = 2 x 14 = 28g

Mass of (CH3)2N2O = 74g

%N = Mass of N / mass of (CH3)2N2O x 100

%N = 28/74 x100

%N = 37.8%

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g