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Sophie [7]
3 years ago
13

4NH3 + 3O2 --> 2N2 + 6H2O

Chemistry
1 answer:
larisa86 [58]3 years ago
8 0

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

 <span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

<span>2.62 mol O2</span><span> (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>

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What are the 4 factors that affect the dissolving of a gas
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7 0
3 years ago
Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2
quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure

V_1 = initial volume of gas

V_2 = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 25^oC=273+25=298K

V_1 = initial volume of gas

V_2 = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(2.5atm)\times (10-1)L

w=-22.5L.atm=-22.5\times 101.3J=-2279.25J

(2) A free isothermal expansion from 1 L to 100 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})

w=-11409.6J

(3) A reversible isothermal expansion from 0.5 L to 4 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})

w=-5151.97J

(4) A reversible isothermal expansion from 0.5 L to 40 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})

w=-10856.8J

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(0.5atm)\times (100-1)L

w=-49.5L.atm=-49.5\times 101.3J=-5014.35J

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

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