1a) What volume of 3.00 mol L−1 HCl in litres is needed to react completely (with nothing left over) with 0.500 L of 0.100 mol L−1 Na2CO3 ?

Answer 1

2HCl + Na2CO3 ==> 2NaCl + H2O + CO2 .. balanced equation

moles Na2CO3 present = 0.500 L x 0.100 mol / L = 0.0500 moles Na2CO3

moles HCl needed (use stoichiometry of balanced eq.): 0.0500 mol Na2CO3 x 2 mol HCl/mol Na2CO3 = 0.1 mol HCl required

volume of 3.00 M HCl: 0.1 mol HCl x 1 L / 3.00 mol = 0.0333 L