A. How many moles of copper equal 8.00 × 109 copper atoms? b. How many moles of calcium equal 102.5 g calcium?
c. How many atoms of lead are present in 5.04 g lead?
d. How many aluminum atoms are present in 2.85 moles of aluminum?
e. What is the mass in grams of 1.08 × 103 moles of fluorine gas?
f. How many molecules of benzene (C6H6) are present in 0.584 g of benzene?
g. What is the mass of 5.09 × 109 atoms of hydrogen gas?
h. How many calcium atoms are present in 0.45 mol Ca3PO4?

Explanation:

a ) one mole = 6.02 x 10²³ atoms

no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

= 1.329 x 10⁻¹⁴ moles .

b ) one mole of calcium = 40 gram

102 .5 g calcium

= 102 .5 / 40 moles

= 2.5625 moles

c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

one mole = 6.02 x 10²³ atoms

2.85 mole = 17.157 x 10²³ atoms .

e )

moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

mass in grams =  .1794 x 10⁻²⁰ x 38

= 6.8172 x 10⁻²⁰ grams

f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

= 2.709 x 10²³ molecules of Ca₃PO₄

no of atoms of Ca = 3 x 2.709 x 10²³

= 8.127 x 10²³ atoms of Ca .

0.30 mol to precipitate all of the ions.

The first step would be to write out the equation as follows:-  

Now we need to balance the reaction:

So from the above balanced reaction, we can see that for every mole of NaCl and every mole of , we need 3 moles of  to precipitate all of the ions.

So,  0.10 mol of KCl and 0.10mol of , we need a total of 0.30mol  to precipitate all of the ions.

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