How much Na2CO3 is needed to neutralize?

Question

How much Na2CO3 is needed to neutralize 200 mL of .05 M Hcl according to the reaction

2HCl + Na2CO3 > 2NaCl + H2O + CO2

the molar mass of Na2Co3 is 106 g/mol, the molar mass of HCl is 36.5 g/mol and the density of the solution is 1.23 g/mL

Answer 1

First we use the balanced equation

2HCl + Na2CO3 > 2NaCl + H2O + CO2

From this two moles of hydrochloric acid will react with one mole of sodium carbonate. The number of moles of hydrochloric acid we have is 0.2*0.05= 0.01moles, which will react with 0.01/2= 0.005 moles of sodium carbonate. The mass of sodium carbonate that is reacted is 0.005*its molar mass= 0.005*106= 0.53grammes

Answer 2

The volume of Na2CO3 that has to be used will depend upon the molarity of Na2CO3 you want to use. Like if you are using 0.05 molar Na2CO3 you have to use 100ml and if its 0.025M you have to use 200ml for neutralisation.
You just have to use this formula
MV/n = M'V'/n'
M and M' are molarities of different solutions
V and V' are the volumes and
n and n' are the no. of moles reacting of the same solutions respectively.

Answer 3

if the .05M means 5 percent solution
then 5 percent of 200 mL is 10mL
10mL of HCl is 1.23x10 =12.3g

2moles of HCL reacts with 1 mole of Na2CO3
i.e. 36,5x2g reacts with 106g
73g of HCl reacts with 106g of Na2CO3
you have 12,3g to react so you need
106x12,3/73=17.9grams of Na2CO3 in solid state.

if the .05M means 0.05 mole of HCl
than you need 106grams x0,05/2

Answer 4

200 x (0.05/1000) x (1/2) moles.

Multiply this by 106 to get the mass.