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defon
3 years ago
8

What is the maxiumum volume of H2 gas at STP that can be produced when 14.0 g of CaH2 and 28.0 g H2O react?

Chemistry
1 answer:
astraxan [27]3 years ago
6 0
You are given the mass of CaH2 which is 14 grams and H2O which is 28 grams. You are required to find the maximum volume of H2 gas at STP. The balanced chemical reaction is CaH2 + 2H2O -> Ca(OH)2 + H2. Note that for every one mole of CaH2, 2 moles of H2O is needed to completely react and produce Ca(OH)2 + H2. So the CaH2 and H2O are at equal proportions. At STP, temperature is at 0°C(273K) and pressure at 1 atm. The molar mass of CaH2 is 42 grams per mole.  

14g CaH2(1mol CaH2/42 grams CaH2)(1mol H2/1mol CaH2) = 0.333 moles H2  

The ideal gas equation is PV=nRT, with R(gas constant) = 0.08206 L-atm/mol-K. Get the equation for volume, we have  

V=nRT/P
V=(0.333molH2)( 0.08206 L-atm/mol-K)(273K)/1atm
<u>V=7.47L</u>
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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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Calculate the volume of the gas, in liters, if 1.35 mol
Alina [70]

Answer:

22.8 L

Explanation:

Step 1: Given data

  • Moles of the gas (n): 1.35 mol
  • Pressure of the gas (P): 1.30 atm
  • Temperature (T): -6°C
  • Ideal gas constant (R): 0.0821 atm.L/mol.K

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = -6 + 273.15 = 267 K

Step 3: Calculate the volume of the gas

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 1.35 mol × (0.0821 atm.L/mol.K) × 267 K / 1.30 atm

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