Question

2Ca(s) + O2(g) → 2CaO(s)

Calculate the mass of calcium oxide that can be prepared from 7.52 g of Ca and 5.01 g of O2.

Answer 1

7.52g of Ca = 0.188 moles
5.01g of CO2 = 0.114 moles
The equation says that 0.188 moles of Ca will use 0.094 moles of CO2, and produce 0.188 moles of CaO = 10.54g

Answer 2

oops!! oobleck misread the problem as CO2 instead of O2.
mols Ca = 7.52/40 = 0.188
mols O2 = 5.01/32 = 0.156
The problem tells you that 0.188 g Ca will use 0.094 mols O2 and produce 0.188 moles of CaO
0.188 mols CaO x 56 g CaO/mol CaO = 10.5 g CaO so the answer comes out the same although the moles O2 are not the same as moles CO2. Some people are just born lucky. BTW, the limiting reagent is Ca and the excess reagent is O2.

Answer 3

Calculate the mass of the quantity of the chemical in bold.
a 2Ca + O2 2CaO
(from 40 g of Ca) ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________
b 4Li + O2 2Li2O (from 30 g of Li)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
c Mg + 2HCl MgCl2 + H2 (from 85 g of MgCl2)