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3 years ago

25 g of NH, is mixed with 4 moles of O, in the given reaction:

Chemistry

2 answers:

worty [1.4K]3 years ago

7 0

Answer:

a) NH₃ is the limiting reactant

b) 44.1 g NO

c) 39.7 g H₂O

Explanation:

The chemical equation is the following:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l)

According to the balanced chemical equation, 4 moles of NH₃ reacts with 5 moles of O₂ to produce 4 moles of NO and 6 moles of H₂O.

a) To determine which is the limiting reactant we have to compare the stoichiometric amounts of reactants (NH₃ and O₂) with the actual amounts.

Stoichiometric amounts:

Molecular weight of NH₃= 14 g/mol + (1 g/mol x 3) = 17 g/mol

Stiochiometric amount of NH₃ = 4 moles NH₃ x 17 g/mol = 68 g NH₃

Stiochiometric amount of O₂ = 5 moles

⇒ Stiochiometric ratio = 68 g NH₃/5 moles O₂

Then, we multiply the stoichiometric ratio by the actual amount of O₂ to calculate the required amount of NH₃:

required NH₃ = 4 moles O₂ x 68 g NH₃/5 moles O₂ = 54.4 g NH₃

Thus, we need 54.4 grams of NH₃ to completely react with 4 moles of O₂, but we only have 25 grams of NH₃. Therefore, NH₃ is the limiting reactant.

b) To calculate the amount of NO formed, we use the limiting reactant (NH₃). According to the equation, 4 moles of NH₃ produce 4 moles of NO.

Molecular weight NO = 14 g/mol + 16 g/mol = 30 g/mol

4 mol NH₃ x 17 g/mol = 68 g NH₃

4 mol NO x 30 g/mol = 120 g NO

⇒ Stiochiometric ratio = 120 g NO/68 g NH₃

Now, we multiply the stoichiometric ratio by the actual amount of NH₃ to calculate the mass of NO formed:

NO formed = 25 g NH₃ x 120 g NO/68 g NH₃ = 44.1 g NO

c) According to the chemical equation, 4 moles of NH₃ (68 g) produce 6 moles of H₂O.

Molecular weight H₂O = (2 x 1 g/mol) + 16 g/mol = 18 g/mol

6 mol H₂O x 18 g/mol = 108 g H₂O

⇒ Stoichiometric ratio = 108 g H₂O/68 g NH₃

Finally, we multiply the stoichiometric ratio by the actual amount of NH₃ (limiting reactant) to obtain the mass of H₂O formed:

H₂O formed = 25 g NH₃ x 108 g H₂O/68 g NH₃ = 39.7 g H₂O

FromTheMoon [43]3 years ago

3 0

Answer:

a. NH3 is limiting reactant.

b. 44g of NO

c. 40g of H2O

Explanation:

Based on the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)

4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.

To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:

a. Moles NH3 -Molar mass. 17.031g/mol-

25g NH3*(1mol/17.031g) = 1.47moles NH3

Moles O2 = 4 moles

For a complete reaction of 4 moles of O2 are required:

4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.

As there are just 1.47 moles, NH3 is limiting reactant

b. Moles NO:

1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO

Mass NO -Molar mass: 30.01g/mol-

1.47mol NO * (30.01g/mol) = 44g of NO

c. Moles H2O:

1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O

Mass H2O -Molar mass: 18.01g/mol-

2.205mol H2O * (18.01g/mol) = 40g of H2O

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Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

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For A: The answer becomes 0.1

For B: The answer becomes 0.4884

Explanation:

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

Rule applied for addition and subtraction:

The least precise number present after the decimal point determines the number of significant figures in the answer.

Rule applied for multiplication and division:

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

For A: (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

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Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

For B: (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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$A+BX\rightarrow AX+B$