Aqueous sulfuric acid (h2so4) reacts with solid sodium hydroxide (naoh) to produce aqeous sodium sulfate (na2so4) and liquid water (h2o). what is the theoretical yield of water formed from the reaction of 5.9 g of sulfuric acid and 6.6 g of sodium hydroxide?

Theoretical yield = 8.55 g

Explanation:

The formula for the calculation of moles is shown below:

For

Mass of water = 5.9 g

Molar mass of water = 98.079 g/mol

The formula for the calculation of moles is shown below:

Thus,

Given: For

Given mass = 6.6 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So,

0.0602 mole of sulfuric acid reacts with 2*0.0602 moles of NaOH

Moles of NaOH = 0.1204 moles

Available moles of = 0.1650 moles

Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of sulfuric acid produces 1 mole of sodium sulfate

So,

0.0602 mole of sulfuric acid produces 0.0602 mole of sodium sulfate

Moles of sodium sulfate = 0.0602 mole

Molar mass of sodium sulfate = 142.04 g/mol

Mass of sodium sulfate = Moles × Molar mass = 0.0602 × 142.04 g = 8.55 g

Theoretical yield = 8.55 g

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm