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777dan777 [17]
3 years ago
13

How many grams of copper (I) chloride can be produced from the reaction of 73.5 g of copper (I) oxide with hydrochloric acid acc

ording to the following reaction? Cu2O + 2 HCl à 2 CuCl + H2O
Chemistry
1 answer:
natulia [17]3 years ago
6 0

The balanced equation for the reaction is as follows

Cu₂O + 2HCl ---> 2CuCl + H₂O

Molar ratio of Cu₂O to CuCl is 1:2

mass of Cu₂O reacted - 73.5 g

Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol

According to the molar ratio,

when 1 mol of  Cu₂O reacts then 2 mol of CuCl is formed

therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed

number of CuCl moles formed - 1.02 mol

mass of CuCl formed - 1.02 mol x 99 g/mol = 101 g

mass of CuCl formed is 101 g

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This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

Q_T=Q_1+Q_2+Q_3+Q_4+Q_5

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\&#10;\\&#10;Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\&#10;\\&#10;Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\&#10;\\&#10;Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\&#10;\\&#10;Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ

Finally, we add them up to get the result:

Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\&#10;\\&#10;Q_T=1.758kJ

Learn more about heating curves: brainly.com/question/10481356

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2 years ago
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