Respuesta :
Answer:
a. AgNO₃ is limiting reactant
b. 26.7g of AgNO₃ would be needed
Explanation:
Based on the reaction:
Cu(s) + 2AgNO₃ → 2Ag(s) + Cu(NO₃)₂
1 mole of Cu reacts with 2 moles of AgNO₃
To find limiting reactant we need to convert mass of each reactant to moles using molar mass:
Moles Cu: Molar mass: 63.546g/mol:
5g * (1mol / 63.546g) = 0.0787moles of Cu
Moles AgNO₃: Molar mass: 169.87g/mol:
20g * (1mol / 169.87g) = 0.1177moles of AgNO₃
For a complete reaction of 0.0787moles of Cu are required:
0.0787moles Cu * (2 moles AgNO₃ / 1mol Cu) = 0.1574 moles AgNO₃
As there are just 0.1177 moles,
AgNO₃ is limiting reactant
b. Would be needed 0.1574 moles of AgNO₃. In grams:
0.1574mol * (169.87g / mol) =
26.7g of AgNO₃ would be needed
a). In the given reaction, the limiting reactant would be:
[tex]AgNO_{3}[/tex]
b). The amount of limiting reactant required for the reaction would be:
[tex]26.7g[/tex]
a). Given the reaction,
[tex]Cu(s) + 2[/tex][tex]AgNO_{3}[/tex] → [tex]2Ag(s) +[/tex][tex]Cu(NO_{3})_{2}[/tex]
As we can see,
Cu's 1 mole displays a reaction with 2 moles of
To find,
Moles of Cu = Mass of Cu/Molar Mass of Cu
[tex]= 5g/63.546g[/tex]
[tex]= 0.0787[/tex] moles
Similarly,
Moles of [tex]AgNO_{3}[/tex] = Mass of [tex]AgNO_{3}[/tex]/Molar Mass of [tex]AgNO_{3}[/tex]
[tex]= 20g/169.87g = 0.1177[/tex] moles
Now,
The entire reaction;
[tex]0.0787 Cu[/tex] × 2 moles of [tex]AgNO_{3}[/tex]/1 mol of Cu
= [tex]0.1574[/tex] moles
Since only 0.01177 moles of [tex]AgNO_{3}[/tex] are present, it is the limiting reactant.
b). To determine the requirement of [tex]AgNO_{3}[/tex],
Moles of [tex]AgNO_{3}[/tex] × molar mass of
[tex]= 0.1574[/tex] × [tex]169.87 g/mol[/tex]
= [tex]26.7g[/tex]
Learn more about "Reactant" here:
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