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kotegsom [21]
3 years ago
8

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

Chemistry
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

a) <u>0.168 moles O2</u>

<u>b) </u> <u>9.50 grams O2</u>

<u>c) 0.01662 kg NO</u>

<u>d)</u>88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen =  6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = <u>0.168 moles O2</u>

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b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol =  <u>9.50 grams O2</u>

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = <u>0.01662 kg NO</u>

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d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

<em>Calculate moles of NH3</em> = 2.51 grams / 17.03 g/mol = 0.147 moles

<em>Calculate moles of O2 </em>= 3.76 grams / 32 g/mol = 0.118 moles

<em>Determine the limiting reactant</em>

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

<em>Calculate moles H2O</em>: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

<em>Calculate mass H2O</em> = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

<em>Calculate % yield</em> = (2.27/2.552)*100 % = <u>88.9 %</u>

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Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

2NH3(aq) + Ag^+(aq) -------->  [Ag(NH3)2]^+(aq)

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Answer:

Option D. Al is above H on the activity series.

Explanation:

The equation for the reaction is given below:

2Al + 6HBr —> 2AlBr₃ + 3H₂

The activity series gives us a background understanding of the reactivity of elements i.e how elements displace other elements when present in solution.

From the activity series of metals, we understood that metal higher in the series will displace those lower in the series.

Considering the equation given above, Al is higher than H in the activity series. Thus, the reaction will proceed as illustrated by the equation.

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A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *
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The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

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According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

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What is the complete ionic equation for NaOHaq plus HCIaq equals H2OI plus NaCIaq?
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Answer:

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Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq)   →  H₂O(l) + Na⁺(aq) + Cl⁻ (aq)

Explanation:

Chemical equation:

NaOH(aq) + HCl(aq)   →  H₂O(l) + NaCl (aq)

Balanced chemical equation:

NaOH(aq) + HCl(aq)   →  H₂O(l) + NaCl (aq)

Ionic equation:

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Net ionic equation:

OH⁻(aq) + H⁺(aq)    →  H₂O(l)

The  Cl⁻(aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

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