Easy chemistry problems, please help!?

Question

Hi, Im takin Chem 201 in college right now, but its been A WHILE since I took chemistry. Im so used to hard stuff, that I completely 4got the easy. Please help with the following questions:

1).

How many moles of molecules and atoms are in a 3.3 g sample of SO3? Use the following atomic masses: S=32, O=16

_____ moles of molecules

_____moles of atoms

2.

A sample of N2O4 contains 2.94 moles of nitrogen. Use the following atomic masses: N =14, O=16.

What is:

The number of moles of N2O4 present =

The mass of the sample =

The number of moles of oxygen in the sample =

The mass of oxygen in the sample =

3).Consider the reaction of 57.0 g of iron as follows:

4Fe + 3O2 2Fe2O3

How many moles of iron react?

How many moles of oxygen must react?

What mass of product forms?

thats all......

Answer 1

To get mols just divide the mass by the molar mass.  

Moles of molecules = Mass/Molar mass of the molecule

=3.3g/(32g + 3 * 16g)

=0.04125

The moles of atoms is easy, there are 4 atoms per each molecule in this case so 4 * 0.04125 = 0.165

2) A little more complex but just use the same formula but check for limiting factors.  For each molecule of N2O4 you have 2 atoms of N and 4 of O.  So they come in a 2/4 ratio, for each 2 atoms of N you have 4 atoms of O

So using the ratio above you know that you have twice as many O's as N's so you must have 0.42 moles of O

Now for the nitrogen

Moles of nitrogen = mass of nitrogen/molar mass of nitrogen

= 2.94g / 14

=0.21 mols

To get the number of moles of N2O4 just use the # of mols of N, remembering that its 2/1 in the equation..

So its 0.21/2 moles of N2O4 = 0.105

Mass of the sample = moles of N2O4 * molar mass of N2O4

=0.105 * (2*14 +  4*16)

=9.66 g

You already have the number of moles of O from before in the solution

Mass of O = mols of O * molar mass of O

=0.42 * 16

=6.72g

3) I think yyou need an = between the O2 and Fe2O3, this is the standard "rust" reaction

Moles of Fe = Mass of Fe / molar mass of Fe

=57/ 55.85

=1.0206 moles

The ratio of Fe to O is 4 to 6 (there are 2 O atoms in that term so its 3*2)

So take the amount of Fe and multipy it by this ratio

=1.0206 *6/4

=1.5309 moles of O

Total mass of the product = mass of the reactants

Total = mass of Fe + Mass of O

= moles Fe * molar mass of Fe  + moles of O * molar mass of O

=57 g + 1.5309 *16

=81.4944

Round to whatever you like