Question

One method of preparing ClO2 is by the reaction of chlorine and sodium chlorite: Cl2 (g) + 2 NaClO2 (s) → 2 ClO2 (g) + 2 NaCl (s) If you allow 15.0 g of NaClO2 to react with 2.00 L of chlorine gas (Cl2) at a pressure of 1.50 atm and at 21 oC, how many grams of ClO2 can be prepared?

Answer 1

To determine the grams of ClO2 that can be prepared from the given reaction, use stoichiometry. First, calculate the moles of NaClO2 and ClO2 using their molar masses and given mass. Finally, calculate the mass of ClO2 using the moles obtained.

Step-by-step explanation:

To determine the grams of ClO2 that can be prepared from the given reaction, we need to use stoichiometry. First, we need to calculate the moles of NaClO2 using its molar mass (74.44 g/mol) and its given mass (15.0 g). moles = mass / molar mass = 15.0 g / 74.44 g/mol = 0.201 mol.

Next, we use the balanced equation to determine the mole ratio between NaClO2 and ClO2, which is 2:2. So, the moles of ClO2 produced are also 0.201 mol.

Finally, we can calculate the mass of ClO2 using its molar mass (67.45 g/mol) and the moles obtained. mass = moles * molar mass = 0.201 mol * 67.45 g/mol = 13.53 g ClO2.

Answer 2

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