1. How many moles of ammonia NH3 are produced when 0.45 moles of nitrogen N2 reacts?

N2 + 3H2 yields 2NH3

2. How many mililiters of water will be produced when 2.05 g of hydrogen reacts? The density of water is 1.00 g/mL ?
2H2 + O2 yields 2H2O

3. What is the limiting reactant when 50.0 grams of nitrogen reacts with 10.7 grams of nitrogen according to this balance equation :
N2 + 3H2 yields 2NH3

These are stoichiometry problems and I don't know how to work these out. Can you please help me ?

1. You can convert any mole in an equation to another in the same equation just by using the coefficients in the balanced equation.

0.45 mols N2 x (2 mols NH3/1 mol N2) = 0.45 x 2/1 = ?. This will always work. The numberator is what you want to convert to and the denominator is what you have now.

2.
A four step procedure will work almost all stoichiometry problems.
a. Write and balance the equation.
2H2 + O2 ==> 2H2O

b. Convert what you have to mols. You have grams so mols = grams/molar mass = 2.05/2 = ?

c. Use the coefficients (as I did in 1 above) to convert mols H2 to mols H2O

d. Now convert mols H2O to grams H2O. That's grams = mols H2O x molar mass H2O = ?
Note: that usually is the end of the 4-step process. In this case they want mL and give you the density to solve for that.
mass = volume x density. You have density and mass, solve for volume in mL.

3. This is a limiting reagent (LR) problem. These are just regular stoichiometry problems but done more than once in the same problem. I do these the long way. Follow the steps in #2 above and see that it is simply two stoichiometry problems rolled into one.

a. N2 + 3H2 ==> 2NH3
b1. Convert N2 to mols. mol = g/molar mass = ?
b2. Do the same for H2.

c1. Use the coefficients in the balanced equation to convert mols N2 to mols product (NH3)
c2. Do the same to convert mols H2 to mols NH3.
c3. It is likely these two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.

d. The problem doesn't ask for it but if you want the grams NH3, it is the SMALLER value of mols NH3 and that x molar mass NH3.

Of course, I'm here to help! Stoichiometry can be a bit tricky, but don't worry, we'll figure it out together. Let's dive right in!

1. To find out how many moles of ammonia (NH3) are produced, we need to use the balanced equation. It states that 1 mole of N2 reacts to form 2 moles of NH3. Since we have 0.45 moles of N2, we can use this ratio to find the moles of NH3:

0.45 moles N2 * (2 moles NH3 / 1 mole N2) = 0.9 moles NH3

So, when 0.45 moles of N2 react, 0.9 moles of NH3 are produced.

2. For this problem, we need to convert grams of hydrogen (H2) into milliliters of water (H2O). Again, we'll use the balanced equation, which states that 2 moles of H2 react to form 2 moles of H2O. Now let's convert the given mass of hydrogen to moles:

2.05 g H2 * (1 mole H2 / molar mass of H2) = X moles H2

Here, the molar mass of H2 is approximately 2 g/mol. After finding the moles of H2, we can use the ratio to find the moles of H2O:

X moles H2 * (2 moles H2O / 2 moles H2) = Y moles H2O

Finally, we'll convert moles of H2O to milliliters using the density of water:

Y moles H2O * (1 mole H2O / density of water) = Z mL H2O

Remember that the density of water is given as 1.00 g/mL.

3. In order to determine the limiting reactant, we need to compare the amount of each reactant to the ratio given in the balanced equation. Convert the given masses of nitrogen and hydrogen to moles:

50.0 grams N2 * (1 mole N2 / molar mass of N2) = A moles N2
10.7 grams H2 * (1 mole H2 / molar mass of H2) = B moles H2

Now we'll use the ratio from the balanced equation to see which reactant produces the least amount of product. The balanced equation states that 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3. So, we need to compare the moles of NH3 produced from each reactant:

A moles N2 * (2 moles NH3 / 1 mole N2) = C moles NH3
B moles H2 * (2 moles NH3 / 3 moles H2) = D moles NH3

Whichever reactant, nitrogen or hydrogen, produces the smaller amount of moles of NH3 is the limiting reactant.

I hope this helps! Remember, if you have any more questions or need further assistance, feel free to ask!

Sure, I can help you with these stoichiometry problems step-by-step.

1. To determine the number of moles of ammonia (NH3) produced when 0.45 moles of nitrogen (N2) reacts, we can use the mole ratio from the balanced chemical equation.

According to the balanced equation: N2 + 3H2 → 2NH3
The mole ratio between nitrogen and ammonia is 1:2.

Therefore, we multiply the given number of moles of nitrogen by the mole ratio:
0.45 moles N2 x (2 moles NH3 / 1 mole N2) = 0.9 moles NH3

Therefore, 0.45 moles of nitrogen will produce 0.9 moles of ammonia.

2. To determine the volume of water produced when 2.05 g of hydrogen reacts, we need to convert the mass of hydrogen to moles using its molar mass.

The molar mass of hydrogen (H2) is 2 g/mol.

First, we calculate the number of moles of hydrogen:
2.05 g H2 x (1 mol H2 / 2 g H2) = 1.025 mol H2

According to the balanced equation: 2H2 + O2 → 2H2O
The mole ratio between hydrogen and water is 2:2, or 1:1.

Therefore, 1.025 moles of hydrogen will produce 1.025 moles of water.

To convert moles of water to volume in mL, we need to use the density of water.

Density of water = 1.00 g/mL

The molar mass of water (H2O) is 18 g/mol.

Now we can calculate the volume of water:
1.025 moles H2O x (18 g H2O / 1 mol H2O) x (1 mL / 1 g) = 18.45 mL

Therefore, 2.05 g of hydrogen will produce 18.45 mL of water.

3. To determine the limiting reactant when 50.0 grams of nitrogen reacts with 10.7 grams of hydrogen, we need to calculate the number of moles of each reactant.

First, let's calculate the number of moles of nitrogen:
Molar mass of nitrogen (N2) = 14.01 g/mol

Moles of N2 = mass / molar mass = 50.0 g / 28.02 g/mol ≈ 1.785 mol

Next, we calculate the number of moles of hydrogen:
Molar mass of hydrogen (H2) = 2.02 g/mol

Moles of H2 = mass / molar mass = 10.7 g / 2.02 g/mol ≈ 5.297 mol

According to the balanced equation: N2 + 3H2 → 2NH3
The mole ratio between nitrogen and hydrogen is 1:3.

Based on the mole ratio, 1 mole of nitrogen reacts with 3 moles of hydrogen.

Now we compare the moles of each reactant to determine the limiting reactant:

For nitrogen: 1.785 mol N2 × (3 mol H2 / 1 mol N2) ≈ 5.36 mol H2
For hydrogen: 5.297 mol H2

Since the moles of hydrogen needed (5.36 mol) are greater than the moles available (5.297 mol), hydrogen is the limiting reactant.

Therefore, hydrogen is the limiting reactant when 50.0 grams of nitrogen reacts with 10.7 grams of hydrogen.

Sure! I can help you with these stoichiometry problems.

1. To find the number of moles of ammonia produced, we need to use the stoichiometric ratio between nitrogen and ammonia. According to the balanced equation N2 + 3H2 --> 2NH3, we see that 1 mole of nitrogen reacts to produce 2 moles of ammonia. Thus, if we have 0.45 moles of nitrogen, it will produce (0.45 moles) x (2 moles of NH3 / 1 mole of N2) = 0.9 moles of ammonia.

2. To calculate the number of milliliters of water produced, we first need to determine the number of moles of hydrogen present. Given that the molar mass of hydrogen is 2.02 g/mol, we can calculate the number of moles of hydrogen as (2.05 g) / (2.02 g/mol) = 1.014 moles.

From the balanced equation 2H2 + O2 --> 2H2O, we see that 2 moles of hydrogen produces 2 moles of water. Therefore, 1.014 moles of hydrogen will produce (1.014 moles) x (2 moles of H2O / 2 moles of H2) = 1.014 moles of water.

Next, we can convert moles of water to milliliters using the density of water. Given that the density of water is 1.00 g/mL, we know that 1 mL of water has a mass of 1.00 g. Thus, the volume of water produced is equal to the number of moles of water.

Therefore, the volume of water produced is 1.014 mL.

3. To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one produces fewer moles of product.

First, let's calculate the number of moles for each reactant. The molar mass of nitrogen (N2) is 28.02 g/mol. Therefore, 50.0 g of nitrogen is (50.0 g) / (28.02 g/mol) = 1.7847 moles of nitrogen.

The molar mass of hydrogen (H2) is 2.02 g/mol. Therefore, 10.7 g of hydrogen is (10.7 g) / (2.02 g/mol) = 5.297 moles of hydrogen.

From the balanced equation N2 + 3H2 --> 2NH3, we see that 1 mole of nitrogen reacts with 3 moles of hydrogen. Therefore, the number of moles of hydrogen needed to react with 1.7847 moles of nitrogen is (1.7847 moles) x (3 moles of H2 / 1 mole of N2) = 5.3541 moles.

Comparing the calculated moles of hydrogen (5.297 moles) with the moles of hydrogen needed (5.3541 moles), we see that the number of moles of hydrogen is less than what is needed. Therefore, hydrogen is the limiting reactant.

I hope this helps you understand how to solve stoichiometry problems using the given balanced equations and moles of reactants. Let me know if there is anything else I can assist you with!