Question

In a combination reaction, 1.61 g of lithium is mixed with 6.70 g of oxygen.

a) Which reactant is present in excess?

(b) How many moles of product are formed?

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

___gLi

___gO2

___gLi2O

Answer 1

First, get reaction

Li(s) + O2(g) --> Li2O(s)

balance

2Li(s) + 1/2O2(g) --> Li2O(s)

4Li(s) + O2(g) --> 2Li2O(s)

now...

mol of Li = mass/MW = 1.61/6.941 = 0.23195 mol of Li

mol of O2 = mass/MW = 6.70/32 = 0.209375 mol of O2

ratios:

4 mol of Li = 2 mol of Li2O

0.23195 mol --> 2/4*0.23195   = 0.115975 mol of LiO2

mol of O2 = 2 mol of Li2O

0.209375 mol of O2 = 2*0.209375 = 0.41875 mol of LiO2

therefore, Li is limiting

b)

moles of product forme d( form lithium) = = 0.115975 mol of LiO2

c)

mass left:

mol of Li left = 0

mol of O2 reacted = 0.23195 /4 = 0.0579875

mol of O2 left = 0.209375 - 0.0579875= 0.151388mol of O2 left

mass of O2 left = mol*MW = 0.151388*32 = 4.84g of O2

mass of L2O formed = mol*MW = 0.115975 *29.8814 = 3.4654 g of Li2O