How much NaCl precipitate would be produced if 2 molarity HCl and 1 molarity sodium carbonate react

this is in the reaction.
Na2Co3 (aq) + 2HCl (aq) yields 2NaCl +H2O+CO2
1 molar Na2CO3 and 2 molar HCl are mixed and allowed to react. All Co2 is allowed to vaporate and then it is boiled. The rest is the precipitate and need to find out theoretically what is produced. thanks.
this would be helpful to be responded as fast as possible. thanks again.

c(Na2CO3)= 1 mol dm^-3 (or M)
c(HCl)= 2 mol dm^-3 (or M)

n(HCl)/n(Na2CO3)= 2/1 (this you can see from the balanced equation) (you need this to calculete if one of the reactants is limiting)

now the problem is you didn't specify what amounts of Na2CO3 and/or HCl is used

you need to know the volumes used or at least one

if you have the volume of let's say Na2CO3 (0,5 dm^3 (500 mL))

then you use n(Na2CO3)/n(NaCl)= 1/2 (again from the balanced equation)

n(NaCl)= 2*n(Na2CO3) (c= n/V; n= m/M)

m(NaCl)/M(NaCl)= 2*c(Na2CO3)*V(Na2CO3)

m(NaCl)= 2*c(Na2CO3)*V(Na2CO3)*M(NaCl)= 2 * 1 mol dm^-3 * 0,5 dm^3 * 58,44 g mol^-1= 58,44 g

;)

ps if you have some additional information post it and i'll try to answer it asap

Since enough is used so that 2 mols of NaCl is produced, 116.88grams is produced