Consider the balanced equation.

Chemistry

2 answers:

Liula [17]3 years ago

7 0

Answer:

The answer is 53.4%

Explanation:

I just took the quiz.

Zolol [24]3 years ago

6 0

The balanced reaction that describes the reaction between hydrogen and nitrogen to produce ammonia is expressed 3H2 + N2 = 2NH3. The yield of the reaction is equal to the actual amount of product divided to the theoretical amount of product multiplied by 100 percent. 26.3 grams of H2 theoretically produces149 grams. The yield is 79 divided by 149 equal to 53.02 percent.

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How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ 63 g