When 1.00 g of copper metal was placed in 50.0mL of .800M silver nitrate solutio

Question

When 1.00 g of copper metal was placed in 50.0mL of .800M silver nitrate solution, 3.21g of silver metal were collected. The unbalanced equation for this reaction is: Cu(s) + AgNO3 (aq) ------> Cu(NO3)2 (aq) + Ag (s)

1. How many moles of copper were present?

2. How many moles of silver were produced?

3. What is the simplest whole number mole ratio between copper and silver?

4. What are the stoichiometric coefficients for each of the reactants and products?

5. How many moles of silver nitrate were initally present?

6. Based on the number of moles of copper and silver nitrate initially present, which one is the limiting reactant?

Explanation / Answer

balanced eqn: Cu(s) + 2AgNO3 (aq) ------> Cu(NO3)2 (aq) + 2Ag (s)

1 gm of copper = 1/63.5 mole of copper

= 0.0157 mole of copper

3.21 gms of silver was produced = 3.21/108 mole ofsilver   

= 0.0297 mole of silver

simplest mole ratio = 0.0297/0.0157

= 1.892

=2

Cu(s) +2AgNO3 (aq) ------> Cu(NO3)2 (aq) +2Ag (s)

stoichometric coefficients:

cu = 1

agno3 = 2

ca(no3)2 = 1

ag=2

5) 2moles of agno3 produce 2 mole of ag

? moles of agno3 produce 0.0297 mole of ag

? = 0.0297 moles of agno3 present in solution

6) silver nitrate is the limiting reagent (no of moles of cu > 2*no of moles of agno3)

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