What is the theoretical yield of n2o5 when 28.0 g of n2 reacts with 68.0 g of o2 in the following equation? 2 n2 + 5 o2 → 2 n2o5

The theoretical yield is 90.7 g.

We know that;

Number of moles of N2= 28.0 g/28 g/mol = 1 moles

Number of moles of oxygen =  68.0 g/32 g/mol = 2.1 moles

From the reaction equation;

2 N2 + 5 O2 → 2 N2O5

2 moles of N2 reacts with 5 moles of O2

1 mole of N2 reacts with 5 × 1/2 = 2.5 moles of O2

This implies that O2 is the limiting reactant.

5 moles of O2 yields 2 moles of N2O5

2.1 moles of O2 yields 2.1 ×  2/5 = 0.84 moles

Theoretical yield of N2O5 = 0.84 moles ×  108 g/mol = 90.7 g

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The theoretical yield of N₂O₅ is 91.8 g.

We have the masses of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.  

M_r:      28.01   32.00    108.01

              2N₂  +  5O₂ ⟶ 2N₂O₅

Mass/g: 28.0      68.0  

Step 2. Calculate the moles of each reactant  

Moles of N₂ = 28.0 g N₂ × (1 mol N₂/28.01 g N₂) = 0.9996 mol N₂

Moles of O₂ = 68.0g O₂ × (1 mol O₂/32.00 g O₂) = 2.125 mol O₂

Step 3. Identify the limiting reactant

Calculate the moles of N₂O₅ we can obtain from each reactant.  

From N₂: Moles of N₂O₅ = 0.9996mol N₂ × (2 mol N₂O₅/2 mol N₂) = 0.9996 mol N₂O₅  

From O₂: Moles of N₂O₅ = 2.125 mol O₂ × (2 mol N₂O₅/5 mol O₂) = 0.8500 mol N₂O₅

The limiting reactant is the one that gives the smaller amount of N₂O₅.

Step 4. Calculate the theoretical yield of N₂O₅.

Mass of N₂O₅ = 0.8500 mol N₂O₅ × (108.01 g N₂O₅/1 mol N₂O₅) = 91.8 g N₂O₅

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

colorless gas, exhibiting a blue glow when placed in an electric field

b) solid

its a type of rock