how many atoms are contained in 2.70g of aluminum provided that 32g of sulphur equals 6.02 × 10^(23)atoms

one atom of an element = 6.02 \times {10}^{23} atom

The mass of one atom of sulphur = 32g

The mass of one atom of aluminium = 27g

so one atom of aluminium = 6.02 \times {10}^{23}

27g of AL = 6.02 \times {10}^{23} atom

2.70g of AL = X atoms

Then you cross multiply ........

and get the answer

.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i believe the answer is a.