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3 years ago

g how many moles of NO are produced if 2.5 mol of O2 react completely with 4.0 mol of NH3 (molar mass of NO

Chemistry

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

2 moles of NO

Explanation:

Our reaction is:

4NH₃ + 5O₂ → 6H₂O + 4NO

4 moles of ammonia react to 5 moles of oxygen in order to produce 6 moles of water and 4 moles of NO

We need to determine the limiting reactant:

Ratio is 4:5, As 4 moles of ammonia react to 5 moles of oxygen and we have 4 moles, we definetely need 5 moles, but the problem is, we only have 2.5 moles of O₂. Oxygen is the limting reagent.

5 moles of O₂ produce 4 moles of NO, according to stoichiometry

2.5 moles of O₂ may produce (2.5 . 4) / 5 = 2 moles of NO

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Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. C2H6 + O2 → CO2 + H2O

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Step 1: Write the unbalanced equation,

C₂H₆  + O₂   → CO₂  + H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

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C₂H₆  + O₂   → 2 CO₂  + 3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

C₂H₆  + 3/2 O₂   → 2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

( C₂H₆  + 3/2 O₂   → 2 CO₂  +  3 H₂O ) × 2

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4 years ago

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3 years ago

A student observed a sample of water in three states of matter. The student should describe the liquid water as a state of matte

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3 years ago

A tank at is filled with of chlorine pentafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal

Ivan

Answer:

- Mole fraction of Chlorine Pentafluoride

= 0.265

- Partial Pressure of Chlorine Pentafluoride

= 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= 0.735

- Partial Pressure of Sulfur Hexafluoride

= 44.53 kPa

Total Pressure exerted by the gases = 60.58 kPa

Explanation:

First of, we calculate the number of moles of each gas present.

Number of moles = (Mass)/(Molar Mass)

For ClF₅

Mass = 4.28 g

Molar Mass = 130.445 g/mol

number of moles of Chlorine Pentafluoride

= (4.28/130.445) = 0.0328 moles

For SF₆

Mass = 13.3 g

Molar Mass = 146.06 g/mol

number of moles of Sulfur Hexafluoride

= (13.3/146.06) = 0.0911 moles

Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.

Using the ideal gas equation

PV = nRT

P = total pressure in the tank = ?

V = volume of the tank = 5.00 L = 0.005 m³

R = molar gas constant = 8.314 J/mol.K

T = temperature of the tank = 20.9°C = 294.05 K

n = total number of moles present = 0.1239 moles

P × 0.005 = (0.1239 × 8.314 × 294.05)

P = 60,580.45 Pa = 60.58 kPa.

- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)

- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)

This is Dalton's law of Partial Pressure.

- Mole fraction of Chlorine Pentafluoride

= (0.0328/0.1239) = 0.265

- Partial Pressure of Chlorine Pentafluoride

= 0.265 × 60.58 = 16.05 kPa

- Mole fraction of Sulfur Hexafluoride

= (0.0911/0.1239) = 0.735

- Partial Pressure of Sulfur Hexafluoride

= 0.735 × 60.58 = 44.53 kPa

Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa

Hope this Helps!!!

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3 years ago