Answer to Question #185529 in General Chemistry for Maddie Dalton

Question

Li2CO3 (s)  +  2HCl (g)  à   H2O (l) +  CO2 (g)  +  2LiCl (aq)  If you combine 5.58 g of lithium carbonate with 0.220 moles of hydrogen chloride,

a)Which is the limiting reactant? 

 

 

 

 

 

b) Which is the excess reactant?

  Please show work for everything Thank you

Answer 1

Li₂CO₃ (s) + 2HCl (g) $\to$ H₂O (l) + CO (g) + 2LiCl (aq)

moles of Li₂CO₃ = 5.58/73.89 = 0.075 moles

According to stoichiometry

For 1 mole of Li₂CO₃ there are 2 moles of HCl

So , for 0.075 moles of Li₂CO₃ there are 0.150 moles of HCl

So we have 0.220 moles of HCl , means HCl is in excess

a ) Li₂CO₃ is the limiting reactant

b) HCl is the excess reactant