In an experiment, 3.25 g of Ali are allowed to react with 3.50 g of O2. 4Li + O2 > 2Li2O. Which reactant is the limiting reagent?

Explanation:

4Li        +       O₂    =     2Li₂O.

4 mole       1 mole

3.25 g Li = 3.25 / 7

= .464 moles

3.5 g of O₂ = 3.5 / 32 mole

= .11 mole

.11 mole of O₂ will react with .44 mole of Li as per the equation above

but Li available is .464 which is in excess .

Hence O₂ is in short supply for given mass of Li .

Hence O₂ is the limiting reagent .

"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"