How many grams of Ag are produced when 36.92 grams of Cu diplaces silver from silver nitrate in the above reaction?
Can you counter check your equation. I understand
Cu + 2 AGNo3 -->Cu(No3)2 + 2 AG, not Cu+AgNo3-->Ag+Cu(NO2)2
Then will shall discuss over the chat link.
Sorry you are correct should read. _Cu+_AgNO3-->_AG+_Cu(NO3)2
So let me work on it, just 10 minutes.
Thank you so much.
Balance the equation: Cu + 2 AGNo3 -->Cu(No3)2 + 2 AG.
Mass of copper = 36.92 grams.
No of moles of copper = its mass/ its molar mass.
Molar mass of copper = 63.5 grams.
No of moles = 36.92/63.5 = 0.5814 moles.
According to the balanced equation, 1 mole of Cu produce 2 moles of Ag, so 0.5814 moles of Cu will produce 0.5814 x 2 moles of Ag, which is 1.1628 moles.
The molar mass of Ag is 107.87 grams.
So 1.1628 mole of Ag equals (107.87 x 1.1628) grams, which is approximately 125.43 grams.
Hope this helps! You can ask any question if you are not satisfied.
Thank you so much for your help. I just could not figure it out. Thank you for clearing it up for me.
Welcome! Thank you for your patience too.
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