What volume of 0.150 M HCl is required for the complete neutralization of 2.00 g of Na2CO3 (sodium carbonate)?

Answer 1

Write the correctly balanced equation:

Na2CO3 + 2HCl ==> 2NaCl + CO2 + H2O

Note the mol ratio in the balanced equation of HCl to Na2CO3, i.e. it is 2 : 1

So, we find moles of Na2CO3, then times 2 to get moles of HCl needed, then convert to volume:

moles Na2CO3 present = 2.00 g x 1 mol/106 g = 0.00943 moles

moles HCl needed = 0.00943 mol x 2 mol HCl/mol Na2CO3 = 0.01886 moles HCl

To find the volume of HCl, we use the fact that it is 0.150 M which = 0.150 mol/L...

(0.150 mol/L)(x L) = 0.01886 moles

x = 0.126 L = 126 ml of HCl needed