Chemistry Problem?

Question

Chlorine is prepared industrially by passing an electric current through brine, a concentrated solution of sodium chloride.
2NaCl + 2H2O → 2NaOH + Cl2 +H2
All three products are commercially valuable. If 1500 g of a brine solution that is 24% NaCl by mass is used, how many grams of each product can be produced?

Answer 1

Atomic weights: Na = 23 Cl = 35.5 H = 1 O = 16 NaCl = 58.5 NaOH = 40 Cl2 = 71 H2 = 2

Let brine be called B

1500gB x 24gNaCl/100gB x 1molNaCl/58.5gNaCl = (1500)(24)/(100)(58.5) = 6.2 mol NaCl (two significant figures)

The 1500gB is given. The first factor comes from the percentage. The grams B cancel, leaving grams NaCl. The second factor comes from the molecular weight. The g NaCl cancel, leaving mol NaCl. I stopped there tempoirarily, because the problem now branches out toward the three products.

6.2molNaCl x 2molNaOH/2molNaCl x 40gNaOH/1molNaOH = (6.2)(2)(40)/(2) = 500 g NaOH

6.2molNaCl x 1molCl2/2molNaCl x 71gCl2/1molCl2 = (6.2)(71)/2 = 880 g Cl2

6.2molNaCl x 1molH2/2molNaCl x 2gH2/1molH2 = (6.2)(2)/(2) = 6.2 g H2.

All to two significant figures.

Answer 2

ok let's start by calculating moles...

mass of NaCl = 1500 g solution x (.24 g NaCl / 1 g solution) = 360 grams

so Moles of NaCl = mass NaCl / molecular weight NaCl
= 360 grams / 58.5 g / mole = 6.15 moles

and moles of water = mass Water / moleculer weight water
= (1500 g - 360 g ) / (18 g/mole) = 63.3 moles

therefore NaCl is the limiting reagent and
6.15 moles NaCl ---> 6.15 moles NaOH and 3.08 moles Cl2 and 3.08 moles H2

mass NaOH = moles x molecular weight = 6.15 x ???
mass Cl2 = moles x molecular weight = 3.08 x ???
mass H2 = moles x molecular weight = 3.08 x ???

you do the rest