Mia K.

asked • 02/09/21

For the following reaction, 118 grams of silver nitrate are allowed to react with 27.1 grams of copper .


What is the maximum amount of copper(II) nitrate that can be formed?

Alex O.

Start out by writing a balanced equation, then you will need to find your limiting reactant. Start by using the given values, convert to mols using the molar mass of the given reactant, next convert from moles of reactant to mols of product, and finally grams of product. Do this for both reactants, the smaller amount in grams (g) of the two will be your answer. Let me know if you would like me to explain further. Here is the stoichiometry for one of the limiting reactant calculations but you will still need to do it using the other reactant to be sure: 118gSilverNitrate*(1mol/MolarMassSilverNitrateGrams)*(molsCopperNitrateFromBalancedEquation/molsSilverNitrateFromBalancedEquation)*(MolarMassCopperNitrateGrams/1mol) = ? Let me know if you would like me to explain further.
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02/09/21

1 Expert Answer

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Hatice B. answered • 02/11/21

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 2 AgNO3+ Cu → Cu(NO3)2 + 2 Ag

118 grams of  AgNO3 is 118/MW=118/169.87=0.6946 moş


27.1 grams of Cu is 27.1/63.546=0.4265


2 moles of Silver nitrate react with 1 mole of copper to produce 1 mole of Copper nitrate


0.6946 moles of Silver Nitrate react with 0.6946/2=0.3473 moles of copper

that gives 0.3473 mole of Copper nitrate

MW of Copper nitrate is 187.56 g/mol

0.3473 mole copper nitrate is 65.1 g copper nitrate


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