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Yakvenalex [24]
3 years ago
14

The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 17

75. In this experiment Lavoisier found that mercury(II) oxide, when heated, decomposed into liquid mercury and an invisible and previously unknown substance: oxygen gas.
Required:
a. Write a balanced chemical equation, including physical state symbols, decomposition of solild mercurytl) oxide (HgO) into liquid mercury and gaseous dioxygen.
b. Suppose 73.0 mL. of dioxygen gas are produced by this reaction, at a temperature of 130.0 °C and pressure of exactly 1 atm. Calculate the mass of mercury() oxide that must have reacted.
Chemistry
1 answer:
yawa3891 [41]3 years ago
5 0

Answer:

a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

b. 0.957 g

Explanation:

Step 1: Write the balanced equation

2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

Step 2: Convert 130.0 °C to Kelvin

We will use the following expression.

K = °C + 273.15

K = 130.0°C + 273.15

K = 403.2 K

Step 3: Calculate the moles of O₂

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K

n = 2.21 × 10⁻³ mol

Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂

The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.

Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO

The molar mass of HgO is 216.59 g/mol.

4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g

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The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4
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Answer:

a

\Delta H^o _{rxn} = -2568.9 \  kJ

b

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c

T_{max}  = 32.4 ^o C

Explanation:

From the question we are told that

 The reaction of cyclobutane and oxygen is

         C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}

ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

Generally ΔH° for this reaction is mathematically represented as

      \Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]

=>  \Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]

=>  \Delta H^o _{rxn} = -2568.9 \  kJ

Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is  mathematically represented as

     H  = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]

=>  H  = [ 4 * 37.1 + 6* 33.6 ]

=>   H  = 350 JK^{-1}

From the question the initial temperature of reactant is  T_i  =  25^oC

Generally the enthalpy change(\Delta H^o _{rxn}) of the reaction is mathematically represented as

 |\Delta H^o _{rxn} |=  H  * (T_{max} -T_i)

  2568.9 =   350  * (T_{max} -25)  

=> \frac{2568.9 }{350}  =  T_{max} - 25

=> T_{max}  = 32.4 ^o C

   

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