CHEM HELP PLS?

Cl₂(g) + 2KOH → KCl + KClO + H₂O
Mole ratio Cl₂ : KOH = 1 : 2
No. of moles of Cl₂ = (0.07589 mol) × (1/2) = 0.03795 mol

For Cl₂ gas :
Pressure, P = 101.325 kPa
Number of moles, n = 0.03795 mol
Gas constant, R = 8.314 J / (mol K)
Temperature, T = 273 K

Gas law : PV = nRT
Volume of Cl₂ gas, V = nRT/P = 0.03795 × 8.314 × 273 / 101.325 L = 0.850 L

Stoichiometry.....

It should be recognized that 0C and 101.3 kPa are the conditions known as STP, standard temperature and pressure. Plus, we all know that 1 mole of any ideal gas at STP occupies a volume of 22.4L.

Cl2(g) + 2KOH(aq) --> KCl(aq) + KOCl(aq) + H2(g)
? ............ 4.25g

4.25g KOH x (1 mol KOH / 56.1g KOH) x (2 mol Cl2 / 1 mol KOH) x (22.4L Cl2 / 1 mol Cl2) = 3.39L

Molar mass of KOH is 56.10 g/mol
Mol KOH in 4.25g = 4.25/56.10 = 0.0758 mol
From the balanced equation
1 mol Cl2 will react with 2 mol KOH
0.0758/2 = 0.0379 mol Cl2 will react with 0.0758 mol KOH
At the STP conditions specified ( 0°C and 101.325 kPa pressure) 1 mol Cl2 has volume = 22.4L
0.0379 mol Cl2 has volume = 0.0379 *22.4 = 0.849L Cl2 gas

If you want a more formal solution:
Use equation
PV = nRT
1*V = 0.0379 * 0.082057*273.15
V = 0.849L Cl2 gas