When 22.0 g NaCl and 21.0 g H, SO4 are mixed and react according to the equation below.

1 answer:

6 0

Taking into account the reaction stoichiometry and limiting reagent, 26.72 grams of Na₂SO₄ are formed when 22 grams of NaOH reacts with 21 grams of H₂SO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCI

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

NaCl: 2 moles
H₂SO₄: 1 mole
Na₂SO₄: 1 mole
HCI: 2 moles

The molar mass of the compounds is:

NaCl: 58.45 g/mole
H₂SO₄: 98 g/mole
Na₂SO₄: 142 g/mole
HCI: 36.45 g/mole

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

NaCl: 2 moles ×58.45 g/mole= 116.9 grams

H₂SO₄: 1 mole ×98 g/mole= 98 grams

Na₂SO₄: 1 mole ×142 g/mole= 142 grams

HCI: 2 moles ×36.45 g/mole= 72.9 grams

<h3>Limiting reagent</h3>

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.

<h3>Limiting reagent in this reaction</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 116.2 grams of NaCl reacts with 98 grams of H₂SO₄, 22 grams of NaCl reacts with how much moles of H₂SO₄?

$mass of H_{2} SO_{4} =\frac{22 grams of NaClx98 grams of H_{2} SO_{4} }{116.2 grams of NaCl}$

<u><em>mass of H₂SO₄= 18.55 grams</em></u>

But 18.55 grams of H₂SO₄ are not available, 21 grams are available. Since you have less moles than you need to react with 22 grams of NaCl, H₂SO₄ will be the limiting reagent.

<h3>Mass of Na₂SO₄ produced</h3>

The following rules of three can be applied, considering the limiting reagent: if by reaction stoichiometry 116.9 grams of NaCl form 142 grams of Na₂SO₄, 22 grams of NaCl form how much mass of Na₂SO₄?

$mass of Na_{2}S O_{4} =\frac{22 grams of NaClx142 grams of Na_{2}S O_{4}}{116.9 grams of NaCl}$