A 500 g block of metal absorbs 5016 J of heat when its temperature changes

For the following reaction, 25.4 grams of sulfur dioxide are allowed to react with 11.6 grams of water . sulfur dioxide(g) water

Gekata [30.6K]

Answer:

The maximum mass of sulfurous acid that can be formed is 32.54 grams.

Explanation:

The balanced reaction is:

SO₂ + H₂O → H₂SO₃

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

SO₂: 1 mole
H₂O: 1 mole
H₂SO3: 1 mole

Being the molar masses of each compound:

SO₂: 64 g/mole
H₂O: 18 g/mole
H₂SO₃: 82 g/mole

Then, by stoichiometry the following quantities of mass participate in the reaction:

SO₂: 1 mole* 64 g/mole= 64 g
H₂O: 1 mole* 18 g/mole= 18 g
H₂SO₃: 1 mole* 82 g/mole=82 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: If 18 grams of water react with 64 grams of sulfur dioxide, how much mass of sulfur dioxide does 11.6 grams of water react with?

$mass of sulfur dioxide=\frac{11.6 grams of water*64 grams of sulfur dioxide}{18 grams of water}$

mass of sulfur dioxide= 41.24 grams

But 41.24 grams of sulfur dioxide are not available, 25.4 grams are available. Since you have less mass than you need to react with 11.6 grams of water, sulfur dioxide will be the limiting reagent.

Then you can apply the following rule of three: if by reaction stoichiometry 64 grams of sulfur dioxide produce 82 grams of sulfurous acid, 25.4 grams of sulfur dioxide, how much mass of sulfurous acid will it produce?

$mass of sulfurous acid=\frac{25.4 grams of sulfur doixide*82 grams of sulfurous acid}{64 grams of sulfur doixide}$

mass of sulfurous acid= 32.54 grams

The maximum mass of sulfurous acid that can be formed is 32.54 grams.

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