Question
A 10.00 g strip of zinc metal is placed in 100. mL of 0.250 M silver nitrate solution. When the reaction that occurs finally ceases, what will be the mass of unreacted zinc metal in the strip? What will be the mass of silver metal that “plates out” of solution (that is, what mass of silver metal is produced by the reaction)?
(1) 9.18 g of unreacted zinc metal remaining; 2.70 g of silver metal produced
(2) 8.36 g of unreacted zinc metal remaining; 2.70 g of silver metal produced
(3) 0.00 g of unreacted zinc metal remaining; 16.5 g of silver metal produced
(4) 8.77 g of unreacted zinc metal remaining: 3.02 g of silver metal produced
(5) 8.36 g of unreacted zinc metal remaining; 3.02 g of silver metal produced
(1) 9.18 g of unreacted zinc metal remaining; 2.70 g of silver metal produced
(2) 8.36 g of unreacted zinc metal remaining; 2.70 g of silver metal produced
(3) 0.00 g of unreacted zinc metal remaining; 16.5 g of silver metal produced
(4) 8.77 g of unreacted zinc metal remaining: 3.02 g of silver metal produced
(5) 8.36 g of unreacted zinc metal remaining; 3.02 g of silver metal produced
Answers
Zn + 2Ag^+ ==> 2Ag(s) + Zn
The problem tells you AgNO3 is the limiting reagent. You have how many mols Ag? That's mols = M x L = 0.250 x 0.1 = 0.0250. Convert mols Ag^+ to mols Zn. That's 1/2 x 0.0250 (from the coefficients in the balanced equation) = 0.0125. So that many mols Zn were used. How many grams is that? That's grams = mols x molar mass = 0.0125 x 65.38 = about 0.817
You had 10g Zn; you used 0.0817. What's left? 10.00 - 0.817 = ?
The problem tells you AgNO3 is the limiting reagent. You have how many mols Ag? That's mols = M x L = 0.250 x 0.1 = 0.0250. Convert mols Ag^+ to mols Zn. That's 1/2 x 0.0250 (from the coefficients in the balanced equation) = 0.0125. So that many mols Zn were used. How many grams is that? That's grams = mols x molar mass = 0.0125 x 65.38 = about 0.817
You had 10g Zn; you used 0.0817. What's left? 10.00 - 0.817 = ?
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