Ammonia is produced when Nitrogen gas reacts with Hydrogen gas, as shown by the equation below:
N2 + 3H2 --> 2NH3
If the percentage yield is 90%, what would be your experimental yield of NH3 if 145.8g of N2 react?
2) 2Mg + O2 --> 2MgO
A) If 50g of Mg reacts with 50g of O2, what is the limiting reactants?
B) How much excess reactant is left over
C) How much product is produced?
To calculate the experimental yield of NH3, follow these steps:
1) Start with the given mass of N2, which is 145.8g.
2) Convert the mass of N2 to moles using the molar mass of N2, which is approximately 28 g/mol.
Mass (N2) = 145.8 g
Molar mass (N2) = 28 g/mol
Moles (N2) = Mass (N2) / Molar mass (N2)
Moles (N2) = 145.8 g / 28 g/mol
Moles (N2) = 5.207 moles
3) Use the balanced equation to determine the stoichiometry between N2 and NH3. From the equation, we can see that 1 mole of N2 reacts to produce 2 moles of NH3.
Moles (NH3) = Moles (N2) x Stoichiometric coefficient (NH3) / Stoichiometric coefficient (N2)
Moles (NH3) = 5.207 moles x 2 / 1
Moles (NH3) = 10.414 moles
4) Calculate the theoretical yield of NH3 by multiplying the moles of NH3 by the molar mass of NH3, which is approximately 17 g/mol.
Theoretical yield (NH3) = Moles (NH3) x Molar mass (NH3)
Theoretical yield (NH3) = 10.414 moles x 17 g/mol
Theoretical yield (NH3) = 176.738 g
5) Finally, calculate the experimental yield of NH3 by multiplying the theoretical yield by the percentage yield.
Experimental yield (NH3) = Theoretical yield (NH3) x Percentage yield / 100
Experimental yield (NH3) = 176.738 g x 90 / 100
Experimental yield (NH3) ≈ 159.064 g
Therefore, the experimental yield of NH3 would be approximately 159.064 g if 145.8 g of N2 reacts.
Now let's move on to the second question about the reaction between 2Mg and O2, forming 2MgO.
A) To determine the limiting reactant, compare the number of moles of Mg and O2, using their masses and molar masses.
Mass (Mg) = 50 g
Molar mass (Mg) = 24 g/mol
Moles (Mg) = Mass (Mg) / Molar mass (Mg)
Moles (Mg) = 50 g / 24 g/mol
Moles (Mg) ≈ 2.083 moles
Mass (O2) = 50 g
Molar mass (O2) = 32 g/mol
Moles (O2) = Mass (O2) / Molar mass (O2)
Moles (O2) = 50 g / 32 g/mol
Moles (O2) ≈ 1.563 moles
From the balanced equation, we can see that the stoichiometric ratio between Mg and O2 is 2:1. Thus, we need twice as many moles of O2 as Mg.
Moles (O2) needed = 2 x Moles (Mg)
Moles (O2) needed = 2 x 2.083 moles
Moles (O2) needed ≈ 4.167 moles
Comparing the moles of O2 needed (4.167 moles) with the actual moles of O2 available (1.563 moles), we see that Mg is the limiting reactant because we have less than half of the required moles of O2.
Therefore, the limiting reactant is O2.
B) To calculate the excess reactant remaining, subtract the moles of O2 needed from the moles of O2 available.
Excess moles (O2) = Moles (O2) available - Moles (O2) needed
Excess moles (O2) = 1.563 moles - 4.167 moles
Excess moles (O2) = -2.604 moles
Since the result is negative, it means there is no excess O2 remaining.
C) To determine the amount of product produced, use the stoichiometric ratio between MgO and O2 from the balanced equation, which is 2:1.
Moles (MgO) = Moles (O2) needed
Moles (MgO) = 4.167 moles
Now, calculate the mass of MgO produced by multiplying the moles of MgO by its molar mass, which is approximately 40 g/mol.
Mass (MgO) = Moles (MgO) x Molar mass (MgO)
Mass (MgO) = 4.167 moles x 40 g/mol
Mass (MgO) = 166.68 g
Therefore, the amount of MgO produced is approximately 166.68 g.
To find the experimental yield of NH3, we need to use the equation and the given percentage yield.
1) Given:
N2 + 3H2 -> 2NH3
Percentage Yield = 90%
Mass of N2 = 145.8g
First, we need to calculate the molar mass of N2:
N2 = 2 * (14.01 g/mol) = 28.02 g/mol
Next, we use the molar ratio between N2 and NH3 to convert the mass of N2 to moles of NH3:
145.8g N2 * (1 mol N2 / 28.02g N2) * (2 mol NH3 / 1 mol N2) = 10.366 mol NH3
Since the percentage yield is 90%, we multiply the calculated moles of NH3 by 0.90 to obtain the experimental yield:
10.366 mol NH3 * 0.90 = 9.33 mol NH3
Finally, we convert the moles of NH3 to grams using its molar mass:
9.33 mol NH3 * (17.03 g NH3 / 1 mol NH3) = 158.589 g NH3
Therefore, the experimental yield of NH3 would be approximately 158.589 grams.
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Now let's move on to the second question regarding the reaction:
2) Given:
2Mg + O2 -> 2MgO
Mass of Mg = 50g
Mass of O2 = 50g
A) To determine the limiting reactant, we need to compare the number of moles of Mg and O2.
First, calculate the molar mass of Mg:
Mg = 24.31 g/mol
Next, convert the masses of Mg and O2 to moles:
Moles of Mg = 50g / (24.31 g/mol) = 2.06 mol
Moles of O2 = 50g / (32 g/mol) = 1.56 mol
The molar ratio in the balanced equation shows that 2 moles of Mg react with 1 mole of O2. Therefore, the limiting reactant is O2 because we have fewer moles of O2 than what is needed for complete reaction.
B) To find out how much excess reactant is left over, we first need to calculate the moles of MgO that can be formed using the limiting reactant.
Since MgO has a 1:1 stoichiometric ratio with O2, the moles of O2 consumed will also determine the moles of MgO formed. Therefore, the moles of MgO are equal to the moles of O2 consumed.
Moles of MgO = Moles of O2 consumed = 1.56 mol
To find the excess reactant, subtract the moles used (1.56 mol O2) from the total moles (2.06 mol Mg) of the excess reactant (Mg).
Excess reactant = Total moles - Moles used
Excess reactant = 2.06 mol - 1.56 mol = 0.5 mol of Mg remains unreacted
C) The amount of product produced can be determined by converting the moles of MgO (which are equal to the moles of O2 consumed) to grams.
Moles of MgO = 1.56 mol
To find the mass of MgO, we need to multiply the moles by its molar mass:
MgO = 1.56 mol * (40.31 g/mol) = 62.87 g
Therefore, approximately 62.87 grams of MgO would be produced.