In a particular reaction 6.80g of dinitrogen trioxide gas (N203) was actually produced by
reacting 8.75g of oxygen gas (O2) with excess nitrogen gas (N2).

a) Balanced chemical equation:

b) Theoretical yield:

c) % yield:

Explanation:

The complete question is:

In a particular reaction 6.80g of dinitrogen trioxide gas (N₂0₃) was actually produced by reacting 8.75g of oxygen gas (O₂) with excess nitrogen gas (N₂)

a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.

b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis

c) Calculate the % yield of the product

a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.

Check the balance:

Atom     Left-handside    Right-hand side

  N              2×2=4                  2×2=4

  O              3×2=6                  2×3=6

b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis

1. Convert 8.75 g of O₂(g) to number of moles

2. Use dimensional analysis to calculate the maximum number of moles of N₂O₃(g) that can be produced

3. Convert to mass in grams

c) Calculate the % yield of the product

Formula:

Substitute and compute:

i'm assuming that the correct answer is indeed the first one/ "a"..