Question

What is the theoretical yield of H 20 that can be obtained from the reaction of 4.5 g H 2 and excess O 2?

2H 2(g) + O2(g) → 2H 20(g)
4.5 g
40.8
80.g
9.0 g
81g

Answer 1

The theoretical yield of H₂O from 4.5 g of H₂ with excess O₂ is calculated to be 20.08 g. This calculation is based on the stoichiometry of the balanced chemical equation for the reaction between hydrogen and oxygen to form water. None of the suggested answers match this calculated value.

Step-by-step explanation:

The theoretical yield of H₂O from the reaction of 4.5 g of H₂ with excess O₂ can be calculated using stoichiometry based on the balanced chemical equation: 2H₂(g) + O₂(g) → 2H₂O(g) First, we need to calculate the number of moles of hydrogen (H₂) using its molar mass (2.02 g/mol). Number of moles of H₂ = mass of H₂ / molar mass of H₂ Number of moles of H₂ = 4.5 g / (2 × 2.02 g/mol) = 1.115 moles of H₂ According to the balanced equation, 2 moles of H₂ produce 2 moles of H₂O. Thus, 1.115 moles of H₂ would theoretically yield 1.115 moles of H₂O. The molar mass of H₂O is approximately 18.02 g/mol, so the mass of H₂O produced can be calculated as follows: Mass of H₂O = moles of H₂O × molar mass of H₂O Mass of H₂O = 1.115 moles × 18.02 g/mol = 20.08 g Therefore, the theoretical yield of H₂O from 4.5 g of H₂ is 20.08 g, which is not one of the suggested answers.