Question

un 15 Complee! If 40.1 g of sulfurous acid reacts with 23.2 g of sodium hydroxide, how many grams of H₂O can be produced? what is the mass of each reactant left after the reaction is over?

Answer 1

The balanced chemical equation for the reaction between sulfurous acid and NaOH is given below,

H2SO3 + 2 NaOH - Na, 503 + 2 H20

Let us now first calculate the number of moles of H2SO3 and NaOH used in the reaction. The molecular weight of H2SO3 and NaOH are, 82 and 40 g/mol.

Number of moles of H2SO3 = Weight of H2SO3 / molecular weight g/mol

= 40.1 g / 82 g/mol = 0.489 moles

Number of moles of NaOH = Weight of NaOH / Molecular weight

= 23.2 g / 40 g/mol = 0.58 moles

It can be seen form balanced chemical equation that 1 mole of H2SO3 requires 2 moles of NaOH.

Therefore, 1 mole of H2SO3 = 2 moles of NaOH

0.489 moles of H2SO3 = (0.489 x 2) = 0.98 moles

As the moles of H2SO3 are in excess, NaOH is the limiting reagent. therefore, yield of H2O will be dependent on the number of moles of NaOH.

It can be seen from the balanced chemical equation, that 2 mole of NaOH gives 2 mole of H2O.

As 0.58 moles of H2SO3 are used, 0.58 moles of H2O will be formed.

Mass of H2O in grams formed = Number of moles of H2O formed x molecular weight

= 0.58 x 18 = 10.44 g

Therefore, grams of of H2O formed will be 10.44 g ,

The excess of reactant (H2SO3) left after completion of reaction = moles of H2SO3 used - moles of H2SO3 consumed

Here number of moles of H2SO3 consumed = Number of moles of NaOH used / 2 = 0.58 / 2 = 0.29 moles

The excess of reactant (H2SO3) left after completion of reaction = 0.489 - 0.29 = 0.2 moles

The mass of excess of H2SO3 is = number of moles of H2SO3 left x molecular weight of H2So3

= 0.2 x 82

= 16.4 g