stoichiometry help please?

2NaOH(aq) + Cl₂(g) → NaCl(aq) + NaOCl(aq) + H₂O (l)


(a)
Mole ratio Cl₂ : NaOCl = 1 : 1
No. of moles of NaOCl produced = (5.73 mol) × (1/1) = 5.73 mol


(b)
Mole ratio NaOH : NaCl = 2 : 1
No. of moles of NaOH reacted = (13.7 mol) × (2/1) = 27.4 mol


(c)
Mole ratio NaOH : H₂O = 2 : 1
No. of moles of H₂O produced = (0.750 mol) × (1/2) = 0.375 mol

Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Mass of H₂O produced = (0.375 mol) × (18.0 g/mol) = 6.75 g

OR:
(0.750 mol NaOH) × (1 mol H₂O / 2 mol NaOH) × (18.0 g H₂O / 1 mol H₂O)
= 6.75 g H₂O


(d)
Mole ratio Cl₂ : NaCl = 1 : 1
No. of moles of Cl₂ reacted = (6.70 mol) × (1/1) = 6.70 mol

Molar mass of Cl₂ = 35.5 × 2 g/mol = 71.0 g/mol
Mass of Cl₂ reacted = (6.70 mol) × (71.0 g/mol) = 476 g

OR:
(6.70 mol NaCl) × (1 mol Cl₂ / 1 mol NaCl) × (71.0 g Cl₂ / 1 mol Cl₂)
= 476 g Cl₂


(e)
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
No. of moles of NaOH reacted = (65.5 g) / (40.0 g/mol) = 1.638 mol

Mole ratio NaOH : NaOCl = 2 : 1
No. of moles of NaOCl produced = (1.638 mol) × (1/2) = 0.819 mol

Molar mass of NaOCl = (23.0 + 16.0 + 35.5) g/mol = 74.5 g/mol
Mass of NaOCl produced = (0.819 mol) × (74.5 g/mol) = 61.0 g

OR:
(65.5 g NaOH) × (1 mol NaOH / 40.0 g NaOH) × (1 mol NaOCl / 2 mol NaOH) × (74.5 g NaOCl / 1 mol NaOCl)
= 61.0 g NaOCl


(f)
No. of moles of Cl₂ = (37.5 g) / (71.0 g/mol) = 0.5282 mol

Mole ratio NaOH : Cl₂ = 2 : 1
No. of moles of NaOH reacted = (0.528 mol) × (2/1) = 1.0564 mol

Mass of NaOH reacted = (1.0564 mol) × (40.0 g/mol) = 42.3 g

OR:
(37.5 g Cl₂) × (1 mol Cl₂ / 71.0 g Cl₂) × (2 mol NaOH / 1 mol Cl₂) × (40.0 g NaOH / 1 mol NaOH)
= 42.3 g

Stoichiometry .....

All of these problems are essentially the same. Start with what is given. Use one or more conversion factors to go from the given units to the unknown units, along with the appropriate numbers. The conversion factors are set up to cancel out the units you don't want and keep the ones you do. When you do the math you will know that it will be correct. Round to the correct number of significant digits.

2NaOH(aq) + Cl2(g) --> NaCl(aq) + NaOCl(aq) + H2O(l)
...................... 5.73 mol .......................? mol
Do in your head: mole ratio is 1:1. Therefore, 5.73 mole of NaOCl

2NaOH(aq) + Cl2(g) --> NaCl(aq) + NaOCl(aq) + H2O(l)
0.750 mol ....................... ....................... .................?g
0.750 mol NaOH x (1 mol H2O / 2 mol NaOH) x (18.0g H2O / 1 mol H2O) = 6.75g H2O

2NaOH(aq) + Cl2(g) --> NaCl(aq) + NaOCl(aq) + H2O(l)
65.5g ......................... ........................? g
65.5g NaOH x (1 mol NaOH / 40.0g NaOH) x (1 mol NaOCl / 2 mol NaOH) x (74.44 g NaOCl / 1 mol NaOCl) = 60.9g NaOCl