Question

To convert from a given quantity of one reactant or product to the quantity of another reactant or product:

First, convert the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadro's number (6.02×1023 particles per mole) to convert number of particles to moles.

Next, convert moles of the given reactant or product to moles of the desired reactant or product using the coefficients of the balanced chemical equation. For example, in the chemical equation

2H2+O2→2H2O

the coefficients tell us that 2 mol of H2 reacts with 1 mol of O2 to produce 2 mol of H2O.

Finally, convert moles of the desired reactant or product back to the desired units. Again, use molar masses to convert from moles to masses, and use Avogadro's number to convert from moles to number of particles.

part A. How many grams of NH3 can be produced from 3.19 mol of N2 and excess H2.

part B. How many grams of H2 are needed to produce 14.94 g of NH3?

part C. How many molecules (not moles) of NH3 are produced from 3.33×10⁻⁴ g of H2?

2. When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is

CH4(g)+O2(g)→CO2(g)+H2O(g)

This type of reaction is referred to as a complete combustion reaction.

Part a. What mass of carbon dioxide is produced from the complete combustion of 2.00×10⁻³ g of methane?

Part b. What mass of water is produced from the complete combustion of 2.00×10⁻³ g of methane?

Part c. What mass of oxygen is needed for the complete combustion of 2.00×10⁻³ g of methane?

3. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

Part a. What is the maximum mass of aluminum chloride that can be formed when reacting 31.0 g of aluminum with 36.0 g of chlorine?

Express your answer to three significant figures and include the appropriate units.

4.

Determine the balanced chemical equation for this reaction.

C8H18(g)+O2(g)→CO2(g)+H2O(g)

Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C8H18, two moles of O2, three moles of CO2, and four moles of H2O.

5. 1.66 g H2 is allowed to react with 10.3 g N2, producing 1.62 g NH3.

Part A

What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

You are given 31.0 g of aluminum and 36.0 g of chlorine gas.

Part B

What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Welcome to Aula Ya Q&A

part A. How many grams of NH₃can be produced from 3.19 mol of N₂and excess H₂.

part B. How many grams of H₂ are needed to produce 14.94 g of NH₃?

part C. How many molecules (not moles) of NH₃ are produced from 3.33×10⁻⁴ g of H₂?

MW of NH₃ = (1x14 g/mol) + (3x1g/mol) = 17 g/mol

MW of N₂ = 2 x 14 g/mol = 28 g/mol

MW of H₂ = 2 x 1 g/mol = 2 g/mol

For part A we need the balanced equation of the reaction:

N₂ + 3H₂ ------> 2NH₃

In the equation we can see that one mol of nitrogen produces 2 mol of ammonia. If we have excess of hydrogen it means that all of the nitrogen reacts so we can establish a relation between the number of mol for nitrogen and the number of mol of product ammonia. You can use the rule of three or conversion factors.

n(NH₃) = 3.19 mol x 2/1 = 6.38 mol.

To convert the mol into grams we multiply by the molecular weight (MW)

g(NH₃) = 6.38 mol x 17 g/mol = 108.46 g

Part B: In this question you see that the limiting reactant is going to be hydrogen because the total amount of ammonia produced from the 3.19 mol is much larger than 14.94 g. So we proceed like these:

First, calculate the number of mol of ammonia: n(NH₃) = 14.94 g / 17 g/mol = 0.88 mol

When going back to the reaction we see that 3 mol of hydrogen produce 2 mol of ammonia. We again can use rule of three or conversion factors.

Second find the number of mol of hydrogen: n(H₂) = 0.88 mol x 3/2 = 1.32 mol

Finally we find the amount of hydrogen by multiplying by the MW:

g(H₂) = 1.32 mol x 2 g/mol = 2.64 g

Part C:

You can find the moles and proceed as before or use another strategy that gives the same result. Using the reaction and the balance coefficients, relate the amount in grams for the reaction:

N₂ + 3H₂ ------> 2NH₃

28 g + 6 g -------> 34 g

So 6 g of H₂ produce 34 g of ammonia. Then 3.33×10⁻⁴ g of H₂ will produce:

g of ammonia = 3.33×10⁻⁴ g x 34 g / 6 g = 18.7 × 10⁻⁴ g

Number of molecules = n(NH₃) x 6.02x10²³ molecules/mol = g(NH₃)/MW x 6.02x10²³ molecules/mol

Number of molecules = 18.7 × 10⁻⁴ g / 17 g/mol x 6.02x10²³ molecules/mol = 6.622 x 10¹⁹ molecules

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2. When methane (CH₄) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is

CH₄(g)+O₂(g)→CO₂(g)+H₂O(g)

This type of reaction is referred to as a complete combustion reaction.

Part a. What mass of carbon dioxide is produced from the complete combustion of 2.00×10⁻³ g of methane?

Part b. What mass of water is produced from the complete combustion of 2.00×10⁻³ g of methane?

Part c. What mass of oxygen is needed for the complete combustion of 2.00×10⁻³ g of methane?

Balanced reaction: CH₄(g)+2O₂(g) → CO₂(g) + 2H₂O(g)

For the three parts we need to calculate the number of mol of methane:

MW(CH₄) = 16 g/mol       MW(CO₂) = 44 g/mol         MW(O₂) = 32 g/mol               MW(H₂O) = 18 g/mol

n(CH₄) = 2.00×10⁻³ g / 16 g/mol = 1.25 x 10^-4 mol

mass of carbon dioxide: n(CO₂) x MW(CO₂)

from the reaction 1 mol of methane produces one mol of CO₂

mass of carbon dioxide: 1.25 x 10^-4 mol x 44 g/mol = 5.5 x 10^-3 g

mass of water: n(H₂O) x MW(H₂O)

from the reaction 1 mol of methane produces two mol of H₂O

mass of water: 2x(1.25 x 10^-4 mol) x 18 g/mol = 4.5 x 10-3 g

mass of oxygen needed: n(O₂) x MW(O₂)

from the reaction we need two mol of oxygen to react with one mol of methane.

mass of oxygen: 2 x (1.25 x 10^-4 mol) x 32 g/mol = 8.0 x 10^-3 g

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3. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl₂(g)→2AlCl₃(s)

Part a. What is the maximum mass of aluminum chloride that can be formed when reacting 31.0 g of aluminum with 36.0 g of chlorine?

Express your answer to three significant figures and include the appropriate units.

For this problem we need to find which of the two reactants the limiting reactant is. We proceed first by calculating the number of mol for each of them:

MW(Al) = 26.982 g/mol

MW(Cl₂) = 2x35.453 = 70.906 g/mol

MW(AlCl₃) = 26.982 g/mol + (3x35.453) = 133.341 g/mol

n(Al) = m(Al) / MW(Al) = 31.0 g / 26.982 g/mol = 1.149 mol

n(Cl₂) = m(Cl₂) / MW(Cl₂) = 36.0 g / 70.906 g/mol = 0.508 mol

We make a first calculation to see how many number of mol of Cl₂ we would need for the 1.149 mol to react:

2 mol of Al react with 3 mol of Cl₂ so for 1.149 mol of Al we need:

n(Cl₂) = 1.149 mol x 3 / 2 = 1.724 mol

We only have 0.508 mol, so the Cl₂ is the limiting reactant and we only can produce the amount of product that can come from the number of mol of Cl₂ we have. So we use that number to calculate the maximum number of mol of AlCl₃ produced. 3 mol of Cl₂ produce 2 mol of AlCl₃ so 0.508 mol will produce:

n(AlCl₃) = 0.508 mol x 2 / 3 = 0.339 mol

mass of AlCl₃ = 0.339 mol x 133.341 g/ mol = 45.203 g

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4. Determine the balanced chemical equation for this reaction.

C₈H₁₈(g)+O₂(g)→CO₂(g)+H₂O(g)

Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C₈H₁₈, two moles of O₂, three moles of CO₂, and four moles of H₂O.

Start with the carbon and put an 8 before CO₂. Find the rest starting with hydrogen and leave oxygen for the last:

C₈H₁₈(g)+25/2O₂(g)→8CO₂(g)+9H₂O(g)

Coeficients are: 1, 25/2, 8, 9

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5. 1.66 g H₂ is allowed to react with 10.3 g N₂, producing 1.62 g NH₃.

Part A: What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

For part A we need the balanced equation of the reaction:

N₂ + 3H₂ ------> 2NH₃

Use the data for molecular weight given above. Start by finding the limiting reactant

n(H₂) = 1.66 g / 2 g/mol = 0.830 mol

n(N₂) = 10.3 g / 28 g/mol = 0.368 mol.

One mol of N₂ reacts with 3 mol of H₂ so how many mol does 0.368 mol need:

n(H₂) = 0.368 mol x 3/1 = 1.104 mol.

The limiting reactant is hydrogen, so the maximum amount of ammonia has to be calculated starting with the number of mol of hydrogen. 3 mol of hydrogen produce 2 mol of ammonia, so we find the amount produced by 0.830 mol:

n(NH₃) = 0.830 mol x 2/3 = 0.553 mol.

The theoretical yield is the maximum amount or theoretical mass of ammonia.

m(NH₃) = 0.553 mol x 17 g/mol = 9.401 g

Part B: What is the percent yield for this reaction under the given conditions?

Since we only obtained 1.62 g NH₃ we can determine the percent yield dividing this amount by the theoretical amount and multiplying by 100:

%yield = (1.62 g/9.401 g) x 100 = 17.232%

Express your answer to three significant figures and include the appropriate units.

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