Question

C. How many grams of Cu react with 5.65 grams of AgNO3?

Answer 1

we need 1.06g of Cu to react with 5.65g of AgNO₃

Step-by-step explanation:

C u +2 AgNO₃ = Cu(NO₃ )₂ + 2 Ag

the mole ratio between Copper and silver nitrate is 1:2 meaning we need 2 moles of silver nitrate for every mole of copper that takes part in the reaction.

the molar mass of Cu is 63.55

the molar mass of AgNO₃ is 107.8682+ 14 + (16x3) = 107.8682+ 14 + 48= 169.87682≈169.88g

for every 63.55 g of Cu reaction with 2(169.88g of AgNO₃)= 339.736g of AgNO₃

if x grams of Cu react with 5.65 grams of AgNO₃

63.55g= 339.736g

x = 5.65g

cross multiply

339.736x = 5.65 x 63.55 = 359.0575

x = 359.0575/339.736 = 1.05687210069≈1.06

we need 1.06g of Cu to react with 5.65g of AgNO₃

Answer 2

Answer: 1.1g of copper

Explanation: We begin by writing the balanced reaction equation.

So we have : Cu(s) + 2AgNO3 (aq) ------> Cu(NO3)2 (aq) + 2Ag (s)

Molar mass of AgNO3 = 107.87 + 14.01 + 3(16.0) = 169.88g/mol, for 2 moles we then have 2 x 169.88 =339.76g.

Atomic mass of Copper = 64g approximately

From the equation the following deductions can be made:

339.76g of AgNO3 reacts with 64g of copper

5.65g of AgNO3 would react with 64/ 339.76 x 5.65 =1.062 approx 1.1g of copper.