Consider the equation for the formation of water. 2H2 O2 Right arrow. 2H2O What is the theoretical yield of H2O if 130 g of H2O is produced from 18 g of H2 and an excess of O2

The theoretical yield of H₂O from 18 g of H₂ is approximately 160.88 g

Explanation:

The given chemical equation of the reaction is given as follows;

2H₂ (g) + O₂ (g) → 2H₂O (l)

The mass of the H₂O produced = 130 g

The mass of H₂ in the reaction = 18 g

Therefore, we have;

Two moles of hydrogen gas, H₂ is added to one mole of oxygen gas, O₂ will produce two moles of water, H₂O

The molar mass of H₂O = 18.01528 g/mol

The molar mass of O₂ = 32 g/mol

The molar mass of H₂ = 2.016 g/mol

The number of moles. n is given by, n = (The mass)/(The molar mass)

The number of moles of H₂ in the reaction = 18 g/(2.016 g/mol) ≈ 8.93 moles

The number of moles of H₂O produced (actual yield) = 130 g/(18.01528 g/mol) ≈ 7.216 moles

Therefore, given that 2 moles of H₂ produces 2 moles of H₂O, the expected  (theoretical) yield of H₂O from 8.93 moles of H₂ is 8.93 moles of H₂O.

The mass of the theoretical yield of H₂O = 8.93 moles × 18.01528 g/mol = 160.8645 grams ≈ 160.88 grams

The theoretical yield of H₂O from 18 grams of H₂ ≈ 160.88 grams

there all right good

explanation:

the answer is b. dilution