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3 years ago

Equal masses (in grams) of hydrogen gas and oxygen gas are reacted to form water. Which substance is limiting?

Chemistry

2 answers:

Gnesinka [82]3 years ago

5 0

Answer:

Option a. Oxygen gas is limiting

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Now, we were told from the question that equal masses of H2 and O2 reacted.

Let the mass of H2 and O2 be 2g each.

Next let us covert this mass to mole as shown below:

Molar Mass of H2 =2x1 = 2g/mol

Assumed mass of H2 = 2g

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of H2 = 2/2 = 1mole

Molar Mass of O2 = 2x16 = 32g/mol

Assumed mass of O2 = 2g

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 2/32 = 0.063mole

Data obtained from our calculations:

Number of mole H2 = 1mole

Number of mole of O2 = 0.063mole

Now to obtain the limiting reactant, do the following:

From the equation,

2moles of H2 required 1mole of O2.

Therefore, 1mole of H2 will require = 1/2 = 0.5mole of 02.

This amount (0.5mole) of O2 obtained is far greater than the amount (i.e 0.063mole) of O2 earlier calculated for. Therefore it is not acceptable.

Now let us turn the tide around.

From the equation,

2moles of H2 required 1mole of O2.

Therefore, Xmol of H2 will require 0.063mole of O2 i.e

Xmol of H2 = 2 x 0.063 = 0.126mol

This amount (0.126mole) of H2 obtained is far lesser than the amount (i.e 1 mole) of H2 earlier calculated for. Therefore it is acceptable. This implies that H2 is the excess reactant and O2 is the limiting reactant.

Sonbull [250]3 years ago

3 0

Answer:

a. Oxygen gas is limiting

Explanation:

hydrogen gas and oxygen gas are reacted to form water

2H₂ + O₂ → 2H₂O

the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂

Given equal masses of H₂ and O₂

assuming 'x' gm for each, no. of moles of each gas =

no. of moles of H₂ = x/2 = 0.5x moles

no.of moles of O₂ = x/32 = 0.031x moles

This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.

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A sample of carbon dioxide occupies a 5.13 dm3 container at STP. What is the volume of the gas at a pressure of 286.5 kPa and a

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Considering the ideal gas law and STP conditions, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

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Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

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.
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In first place, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide, 5.13 L (5.13 dm³= 5.13 L, being 1 dm³= 1 L) are occupied by how many moles of carbon dioxide?

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Then, you know:

P= 286.5 kPa= 2.8275352 atm (being 1 kPa= 0.00986923 atm)
V= ?
T= 12.9 C= 285.9 K (being 0°C= 273 K)
R= 0.082 $\frac{atmL}{mol K}$
n= 0.229 moles

Replacing in the ideal gas law:

2.8275352 atm× V = 0.229 moles×0.082 $\frac{atmL}{mol K}$ × 285.9 K

Solving:

V= (0.229 moles×0.082 $\frac{atmL}{mol K}$ × 285.9 K)÷ 2.8275352 atm

Finally, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.

Learn more about

STP conditions:

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the ideal gas law:

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