Respuesta :
Upon the complete reaction of 25.5 grams of silver nitrate with excess copper(ii) chloride, 0.075 moles of silver chloride will be formed.
When silver nitrate reacts with copper(II) chloride, silver chloride and copper(II) nitrate are produced. The balanced equation for this reaction is:
[tex]2AgNO_{3(aq)} +CuCl_{2(s)}[/tex] ⇒ [tex]2AgCl_{(s)} +Cu(NO_{3})x_{2(aq)}[/tex]
Using the formula to find the molar mass M = m / n, get the number of moles of silver nitrate when its mass is 25.5 grams.
M = m / n
where M = molar mass
m = mass of a substance
n = number of moles of the substance
M = m / n
169.87 g/mol = 25.5 grams / n mol
n = 0.1501147937 mol
From the balanced equation of the reaction, 2 moles of silver nitrate reacted with 1 mole of copper(ii) chloride will produce 2 moles of silver chloride and 1 mole copper(II) nitrate.
If 25.5 grams of silver nitrate is equal to 0.1501147937 mol, and for every 2 moles of it will produce 1 mole of silver chloride, then the reaction will produce 0.075 mol silver chloride.
n = 0.1501147937 mol / 2
n = 0.07505739683 mol
To learn more about balanced chemical equations: https://brainly.com/question/26694427
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