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.

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To answer these questions, we will follow a step-by-step approach:

1) Write a balanced equation.

To write a balanced equation, we need to determine the chemical formulas of the reactants and the product. Nitrogen is represented by the symbol N, phosphorus by P, and trinitrogen diphosphide by N2P3.

The balanced equation is as follows:

3N2 + 5P → 2N2P3

2) Determine which reactant is limiting and which is in excess.

To determine the limiting reactant, we need to compare the stoichiometry of the reactants based on the balanced equation.

- Nitrogen: 10.0 grams
- Phosphorus: 10.0 grams

By using the molar mass of each element, we can convert grams to moles:

- Nitrogen: 10.0 g / 28.02 g/mol (molar mass of N) = 0.357 mol
- Phosphorus: 10.0 g / 30.97 g/mol (molar mass of P) = 0.323 mol

From the balanced equation, we can see that the mole ratio of N to P is 3:5. So, for every 5 moles of P, we need 3 moles of N.

Comparing the moles of each reactant, we see that we have more than enough N to react with P. Therefore, Nitrogen is in excess, and Phosphorus is the limiting reactant.

3) Calculate the theoretical yield of trinitrogen diphosphide.

Using the stoichiometry from the balanced equation, we can determine the number of moles of trinitrogen diphosphide formed.

From the balanced equation, the mole ratio of N2P3 to P is 2:5. Therefore, the calculated moles of N2P3 formed can be calculated as:

0.323 mol P (1 mol N2P3 / 5 mol P) = 0.065 mol N2P3

Now, we can convert the moles of N2P3 to grams using its molar mass:

0.065 mol N2P3 * (98.89 g/mol) = 6.43 grams of N2P3 (theoretical yield)

So, theoretically, 6.43 grams of trinitrogen diphosphide can be produced.

4) Determine if it is possible to make 20.0 grams of trinitrogen diphosphide.

Since the theoretical yield is 6.43 grams, it is not possible to produce 20.0 grams of trinitrogen diphosphide because it exceeds the theoretical limit.

5) Calculate the percent yield if 15.8 grams of trinitrogen diphosphide is obtained.

The percent yield can be calculated using the following formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Given:
- Actual Yield = 15.8 grams
- Theoretical Yield = 6.43 grams

Plugging into the formula:

Percent Yield = (15.8 g / 6.43 g) * 100 = 245.4%

Therefore, the percent yield is 245.4%. Note that a percent yield greater than 100% indicates that the actual yield is greater than the theoretical yield, which can be due to experimental errors or impurities.