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Lena [83]
2 years ago
9

Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 spa

ce O subscript 2 space (g )space rightwards arrow 3 space C O subscript 2 space (g )space plus space 4 space H subscript 2 O space (g )How many liters of carbon dioxide are produced at STP when 44 g of C3H8 completely reacts with oxygen
Chemistry
1 answer:
Misha Larkins [42]2 years ago
4 0

Explanation:

The balanced chemical equation of the reaction is:

C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of CO_2 gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

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125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure
dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL   (as 1 mL = 0.001 L) = 0.125 L

Temperature = 70^{o}C = (70 + 273) K = 343 K

Pressure = 125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L

Hence, volume of the gas at STP is 22.53 L.

5 0
2 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
What is a cone of depression
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A cone of depression occurs in an aquifer when groundwater is pumped from a well. In an unconfined aquifer, this is an actual depression of the water levels. In confined aquifers, the cone of depression is a reduction in the pressure head surrounding the pumped well.

8 0
3 years ago
Read 2 more answers
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
Steel cranes can carry heavy steel beams thousands of feet high during skyscraper construction. This is an example of:
beks73 [17]

Hello!

The steel cranes lifting thousands of feet high during skyscraper construction is an example of the evolution of building techniques.

I hope it helps!

4 0
3 years ago
Read 2 more answers
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