Which substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes? a)CH4 b)H2S c)CO2 d)NaCl

Chemistry

1 answer:

schepotkina [342]3 years ago

8 0

Answer:

$H_2S$

Explanation:

Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.

Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:

$\Delta H^o = \frac{58.16 kJ}{3.11 mol} = 18.7 kJ/mol$

Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of $H_2S$ is 18.7 kJ/mol

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3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O

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Considering the reaction stoichiometry, the amount of moles of NO that is produced when 27 moles of HNO₃ react is 6.75 moles.

<h3 /><h3>Reaction stoichiometry</h3>

The balanced reaction is:

3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 moles
NO: 2 moles
H₂O: 4 moles

<h3>Amount of moles of NO produced</h3>

It is possible to determine the the amount of moles of NO produced by a rule of three: if by stoichiometry 8 moles of HNO₃ produce 2 moles of NO, if 27 moles of HNO₃ react how many moles of NO will be formed?

$amount of moles of NO=\frac{27 moles of HNO_{3} x 2 moles of NO}{8 moles of HNO_{3}}$