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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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sweet [91]

So let's convert this amount of mL to grams:

\frac{13.6g}{1mL}*1.2mL=16.32g

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

\frac{1mole}{200.59g}*16.32g=8.135*10^{-2}mol

Then we need to convert moles to atoms using Avogadro's number:

\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms

So now we know that in 1.2 mL of liquid mercury, there are 4.90*10^{22}atoms present.

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