Answer:
Limiting reactant: O₂
Excess reactant: NO
Amount of NO₂ produced: 28.75 g
Explanation:
The balanced equation for the reaction between O₂ and NO to produce NO₂ is the following:
O₂(g) + 2NO(g) → 2NO₂(g)
According to the equation, 1 mol of O₂ reacts with 2 moles of NO to produce 2 moles of NO₂.
We first convert the moles of each compound to mass by using the molecular weight (MW):
MW(O₂) = 16 g/mol x 2 = 32 g/mol
⇒ mass O₂ = 1 mol O₂ x 32 g/mol O₂ = 32 g
MW(NO) = 14 g/mol + 16 g/mol = 30 g/mol
⇒ mass NO = 2 mol NO x 30 g/mol NO = 60 g
MW(NO₂) = 14 g/mol + (2x 16 g/mol) = 46 g/mol
⇒ mass NO = 2 mol NO₂ x 46 g/mol NO₂ = 92 g
For the reactants, the stoichiometric ratio is 32 g O₂/60 g NO. Thus, the mass of O₂ required to completely react with 20.00 g of NO is:
mass of O₂ required = 20.00 g NO x 32 g O₂/60 g NO = 10.67 g O₂
We need 10.67 g of O₂ and we have only 10.00 g, so the limiting reactant is O₂. Thus, the excess reactant is NO.
With the limiting reactant, we calculate the amount of product produced (NO₂). For this, we consider the stoichiometric ratio 92 g NO₂/32 g O₂:
mass of NO₂ produced = 10.00 g O₂ x 92 g NO₂/32 g O₂ = 28.75 g NO₂
; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.
= 0.5M