Question

In a combination reaction, 1.51 g of lithium is mixed with 6.71 g of oxygen.

(a) Which reactant is present in excess?

	lithium.
	oxygen.

(b) How many moles of product are formed?

mol

(c) After the reaction, how many grams of each reactant and product are present?

____g Li

____g O₂

____ g Li₂O

Answer 1

    4Li   + O2 -------------> 2Li2O

no of moles of Li   = W/G.M.Wt

                              = 1.51/7   = 0.22 moles

no of moles of O2   = W/G.M.Wt

                                = 6.71/32   = 0.21 moles

    4Li   + O2 -------------> 2Li2O

4 moles of Li react with 1 moles of O2

0.22 moles of Li react with = 1*0.22/4   = 0.055 moles of O2 is required

   O2 is excess reactant

Li is limting reactant

4Li   + O2 -------------> 2Li2O

4 moles of Li react with O2 to gives 2 moles of Li2O

0.22 moles of Li react with O2 to gives = 2*0.22/4   = 0.11 moles of Li2O

mass of Li2O   = 0.11 * 30   = 3.3g of Li2O

excess react = 0.21-0.055   = 0.155 moles of O2

mass of O2   = 0.155*32   = 4.96g

mass of Li     = 0.22*7   = 1.54 g