Question

In a particular reaction 6.80g of dinitrogen trioxide gas (N203) was actually produced by

Answer 1

Answer:

a) Balanced chemical equation:

      $2N_2(g)+3O_2(g)\longrightarrow 2N_2O_3(g)$

b) Theoretical yield:

• 27.4 g of N₂O₃

c) % yield:

• 24.5%

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

Solution

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

      $2N_2(g)+3O_2(g)\longrightarrow 2N_2O_3(g)$

Check the balance:

Atom     Left-handside    Right-hand side

  N              2×2=4                  2×2=4

  O              3×2=6                  2×3=6

• Mole ratio: it is the ratio of the coefficients of the balanced equation

        $\dfrac{2molN_2O_3}{3molO_2}$

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

• number of moles = mass in grams / molar mass
• molar mass of O₂ = 15.999g/mol
• number of moles = 8.75g / 15.999 g/mol = 0.5469 mol O₂

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

      $\dfrac{2molN_2O_3}{3molO_2} \times 0.5469molO_2=0.3646molN_2O_3$

3. Convert to mass in grams

• mass = number of moles × molar mass
• molar mass of N₂O3 = 76.01g/mol
• mass = 0.3646mol × 76.01g/mol = 27.7g N₂O3

c) Calculate the % yield of the product

Formula:

• %yield = (actual yield/theoretical yield)×100

Substitute and compute:

• % yield = (6.80g/27.7g)×100 = 24.5%