The reaction of 2.30g N2 with excess H2 produces 2.00g NH3. The percent yield of this reaaction is?
percent yield = (actual yield/theoretical yield)*100 = ?
2.00 g is given in the problem as the actual yield. You must calculate the theoretical yield. The easy way to do that is with a planned approach to stoichiometry problems. Here is a step by step procedure.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained) with the theoretical yield (the maximum possible amount of product that can be obtained based on stoichiometry).
In this case, we are given the mass of nitrogen gas (N2) and the mass of ammonia (NH3) produced. We'll follow these steps to find the percent yield:
1. Determine the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
2. Calculate the molar mass of N2 and NH3:
Molar mass of N2 = 2(14.01 g/mol) = 28.02 g/mol
Molar mass of NH3 = 2(1.01 g/mol) + (14.01 g/mol) = 17.03 g/mol
3. Convert the given masses of N2 and NH3 to moles:
Moles of N2 = 2.30 g / 28.02 g/mol = 0.082 mol (rounded to three decimal places)
Moles of NH3 = 2.00 g / 17.03 g/mol = 0.117 mol (rounded to three decimal places)
4. Determine the limiting reagent:
Since we have an excess of hydrogen gas (H2), nitrogen gas (N2) is the limiting reagent.
5. Use stoichiometry to find the theoretical yield of NH3:
From the balanced equation, we know that 1 mole of N2 produces 2 moles of NH3.
Therefore, in this reaction, 0.082 mol of N2 would produce 2 * 0.082 mol = 0.164 mol of NH3.
6. Calculate the theoretical mass of NH3:
Mass of NH3 = moles of NH3 * molar mass of NH3
= 0.164 mol * 17.03 g/mol = 2.79 g (rounded to three decimal places)
7. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (2.00 g / 2.79 g) * 100 = 71.68% (rounded to two decimal places)
Therefore, the percent yield of this reaction is approximately 71.68%.