This is the question
10.0 grams of nitrogen is reacted with 10.0 grams of the element phosphorus under certain conditions to yield trinitrogen diphosphide. Answer these questions.
1). Write a balanced equation.
2). which reactant is limiting? In excess?
3). Theoretically, how many grams of trinitrogen diphosphide can be produced?
4). Is it possible to make 20.0 grams of trinitrogen diphosphide?
5). If 15.8 grams of trinitrogen diphosphide is found during an experiment, what is the percent yield.
1) To write a balanced equation, we need to determine the chemical formula for each compound involved in the reaction. The reactants are nitrogen (N2) and phosphorus (P4), and the product is trinitrogen diphosphide (N2P3).
The balanced equation is:
3N2 + 2P4 → 2N2P3
2) To determine the limiting reactant, we compare the moles of each reactant present to their stoichiometric coefficients in the balanced equation.
First, let's convert the mass of nitrogen and phosphorus to moles using their molar masses. The molar mass of nitrogen is 14.01 g/mol, and the molar mass of phosphorus is 30.97 g/mol.
For nitrogen:
moles = mass / molar mass
moles = 10.0 g / 14.01 g/mol = 0.713 mol
For phosphorus:
moles = mass / molar mass
moles = 10.0 g / 30.97 g/mol = 0.323 mol
Now, let's compare the moles of each reactant to their stoichiometric coefficients. From the balanced equation, we can see that the ratio of N2 to P4 is 3:2.
The moles ratio of N2:P4 in the given amounts is:
0.713 mol N2 / 0.323 mol P4 = 2.21
Since the ratio is less than the stoichiometric ratio of 3:2, phosphorus is the limiting reactant, and nitrogen is in excess.
3) To find the theoretical yield of trinitrogen diphosphide, we need to calculate how many moles of the limiting reactant, phosphorus (P4), we have and convert it to moles of trinitrogen diphosphide (N2P3).
From the balanced equation, we can see that 2 moles of N2P3 are produced for every 2 moles of P4. Therefore, the moles of N2P3 produced will be equal to the moles of P4.
moles of N2P3 = 0.323 mol P4
To convert moles of N2P3 to grams, we can use the molar mass of N2P3. The molar mass of N2P3 is 166.92 g/mol.
mass of N2P3 = moles of N2P3 × molar mass of N2P3
mass of N2P3 = 0.323 mol × 166.92 g/mol = 53.89 g
Theoretically, 53.89 grams of trinitrogen diphosphide can be produced.
4) To determine if it is possible to make 20.0 grams of trinitrogen diphosphide, we compare the given mass to the theoretical yield obtained in the previous step.
Since the theoretical yield is 53.89 grams, which is greater than 20.0 grams, it is possible to make 20.0 grams of trinitrogen diphosphide.
5) To calculate the percent yield, we need both the actual yield (15.8 grams) and the theoretical yield (53.89 grams) of trinitrogen diphosphide.
percent yield = (actual yield / theoretical yield) × 100
percent yield = (15.8 g / 53.89 g) × 100 = 29.31%
The percent yield is 29.31%.