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densk [106]
3 years ago
14

Hydrobromic acid dissolves solid iron according to the following reaction:

Chemistry
1 answer:
Alisiya [41]3 years ago
4 0

Answer:

8.1g

0.1g

Explanation:

The reaction expression is given as:

 

      Fe   +   2HBr   →  FeBr₂    +    H₂

Mass of pure iron given  = 2.8g

A. Mass of HBr needed to dissolve a padlock of the mass;

To solve this problem, we need to use the mole concept.

 Convert mass of the known iron to the number of moles.

  Number of moles = \frac{mass}{molar mass}  

  Molar mass  = 56g/mol  

   Number of moles of iron  = \frac{2.8}{56}   = 0.05mole

     

 1 mole of Fe will react with 2 mole of HBr

 0.05mole of Fe will react with 0.05 x 2  = 0.1mole of HBr

 Mass of HBr  = number of moles x molar mass

   Molar mass of HBr  = 1 + 80   =  81g/mol

 Mass of HBr  = 0.1 x 81  = 8.1g

B. What mass of H2 would be produced by the complete reaction of the iron bar

    Since:

         1 mole of Fe will produce 2 mole of hydrogen gas

        0.05mole of Fe will produce 2 x 0.05mole  = 0.1mole of hydrogen gas

 Mass of hydrogen gas  = number of moles x molar mass

                                          = 0.1 x 1

                                          = 0.1g

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Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing the gas at a pressure of 10
faust18 [17]

Answer:

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Pressure = 0.6907 bar.

dT/dt = - 1,151 K/s.

Explanation:

The first thing to do here is to write out the equation for mass balance as given below:

dN/dt = N -------------------------------------------------------------------------------------------(1).

N = P/T, then, substitute the values given in the question into:

d[p/T]/ dt = [- 4.5/28 × 8.314]/0.15 = - 8.9 × 10⁻⁵ bar/K.s.

Thus, there is the need to integrate, Integrate [p/T]f = 10/320 - 8.9 × 10⁻⁵ bar/K.s. ------------------------------------(2).

NB; fT = final temperature, fP = final pressure and iT = initial temperature.

Also, [ fT]³⁰/₈.₃₁₄/ [fP] = [iT]³⁰/₈.₃₁₄/ Pi] = [ 320]³⁰/₈.₃₁₄/ 10.

Therefore, [fT]³⁰/₈.₃₁₄ = 109.52 × 10⁶.

Final temperature=  [fP]³⁰/₈.₃₁₄ × 169.05.

Note that fP/ [fP]³⁰/₈.₃₁₄ × 169.05 = 10/320 - 8.9 × 10⁻⁵.

Therefore, [fP]¹ ⁻ ³⁰/₈.₃₁₄ = 0.7651.

Hence, Final temperature = 152.57K,

Pressure = 0.6907 bar

dT/ dt = N[RT]² / Cv . PV.

R = 30 - 8.314 = 21.86 J/mol K.

Then, the rate of change of the gas temperature at this time = dT/dt = - 1,151 K/s.

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