Answer:
5.59 g of ClO₂
Solution:
First calculate moles of Cl₂:
According to ideal gas equation,
P V = n R T
Or,
n = P V / R T
Putting values,
n = (1.5 atm × 2.0 L) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 294 K)
n = 0.124 mol of Cl₂
According to equation,
1 mole Cl₂ required = 2 moles of NaClO₂
So,
0.124 mole Cl₂ will require = X moles of NaClO₂
Solving for X,
X = (0.124 mol × 2 mol) ÷ 1 mol
X = 0.248 mol of NaClO₂
And given mass of NaClO₂ equals followinf moles,
n = 15 g ÷ 90.45 g.mol⁻¹
n = 0.1658 mol of NaClO₂
It means we are given with less amount of NaClO₂, hence it is limiting reagent and will control the amount of product formed.
So, According to equation
2 moles of NaClO₂ produced = 67.45 g of ClO₂
So,
0.1658 moles of NaClO₂ will produce = X g of ClO₂
Solving for X,
X = (67.45 g × 0.1658 mol) ÷ 2 mol
X = 5.59 g of ClO₂
