Answer:
Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g
Explanation:
SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.
the balanced chemical equation is as follows
Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O
Moles of Na₂SO₃ = ![\frac{Mass}{Molecular mass} =\frac{25}{126} = 0.198](https://tex.z-dn.net/?f=%5Cfrac%7BMass%7D%7BMolecular%20mass%7D%20%3D%5Cfrac%7B25%7D%7B126%7D%20%3D%200.198)
Moles of HCl = ![\frac{mass}{molecular mass}=\frac{22}{36.5}= 0.6](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolecular%20mass%7D%3D%5Cfrac%7B22%7D%7B36.5%7D%3D%200.6)
using mole ratio method to find limiting reagent
For sodium sulfite ![\frac{mole}{stoichiometry} = \frac{0.198}{1}= 0.198](https://tex.z-dn.net/?f=%5Cfrac%7Bmole%7D%7Bstoichiometry%7D%20%20%3D%20%5Cfrac%7B0.198%7D%7B1%7D%3D%200.198)
for HCl ![\frac{mole}{stoichiometry} = \frac{0.6}{2}= 0.3](https://tex.z-dn.net/?f=%5Cfrac%7Bmole%7D%7Bstoichiometry%7D%20%20%3D%20%5Cfrac%7B0.6%7D%7B2%7D%3D%200.3)
since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction
1 mole of Na₂SO₃ produce 1 mole of SO₂
0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂
∴ Mass of SO₂ produce = mole x molar mass of SO₂
= 0.198 x 64
= 12.672 g