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SSSSS [86.1K]
3 years ago
9

Nitrogen dioxide, a major air pollutant, can be produced by the combustion of nitrogen oxide as shown.

Chemistry
2 answers:
oksian1 [2.3K]3 years ago
8 0
Percent yield is calculated by obtaining the ratio of the actual yield and the theoretical yield times 100. 

Actual yield = 1500 kg

Theoretical yield = 1500 ( 1 / 30.00 ) ( 2 / 2 ) ( 46.006 / 1 ) = 2300.3 kg

Percent yield = 1500 / <span>2300.3</span> x 100 =<span> 65.21 %

Therefore, the correct answer is the last option. Hope this answers the question.</span>
xxTIMURxx [149]3 years ago
8 0

Answer is: the percent yield is 65.2%.

Balanced chemical reaction: 2NO + O₂ → 2NO₂.

m(NO) = 1500 kg; mass of nitrog oxide.

From balanced chemical reaction: n(NO) : n(NO₂) = 2 : 2 (1 : 1).

m(NO) : M(NO) = m(NO₂) : M(NO₂).

1500 kg : 30 g/mol = m(NO₂) : 46 g/mol.

m(NO₂) = 1500 kg · 46 g/mol ÷ 30 g/mol.

m(NO₂) = 2300 kg.

the percent yield = 1500 kg ÷ 2300 kg · 100%.

the percent yield = 65.21%.

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What is the density of ammonia gas, nh3, at 31°c and 751 mmhg?
solmaris [256]
First, we will use the general gas formula to get the number of moles.
PV = nRT where:
P is the pressure of gas = 751 mmHg = 100125.096375 Pascal
V is the volume = 1 liter = 0.001 m^3
n is the number of moles we want to calculate
R is the gas constant = <span>8.314 J/(K. </span>mol<span>)
T is the temperature = 31 degrees celcius = </span>304.15 degree kelvin

Substitute in the above equation to get the number of moles as follows:
100125.096375 * 0.001 = n * 8.314 * 304.15
n = 0.039595 moles

Now, we will use the number of moles to get the mass as follows:
number of moles = mass / molar mass
mass = number of moles * molar mass
number of moles = 0.039595 moles
molar mass of ammonia (NH3) = 14 + 3(1) = 17 grams
Substitute to get the mass as follows:
mass = 0.039595 * 17 = 0.673122 grams

Last step is to get the density as follows:
density = mass / volume
mass = 0.673122 grams
volume = 1 liter
density = 0.673122 / 1 = 0.673122 grams/liter = <span>0.000675 kg/L</span>



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