Chemistry homework

Question

What mass of silver can be produced by the reaction of 2.00 g of zinc metal with 6.00 g of silver (I) nitrate?

Answer 1

Zn(s) + 2AgNO₃(aq) ==> Zn(NO₃)₂(aq) + 2Ag(s) ... balanced redox equation

Since mass of both reactants is given, we must first find which, if any reactant, is limiting. An easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever is less is the limiting reactant.

For Zn we have ... 2.00 g Zn x 1 mol Zn / 65.4 g = 0.0306 mols Zn (÷1->0.0306)

For AgNO₃ we have ... 6.00 g AgNO₃ x 1 mol / 170 g = 0.0353 mols AgNO₃ (÷2->0.0176)

Since 0.0176 is less than 0.0306, AgNO₃ is limiting and will determine the amount of product formed.

Using moles of AgNO₃, we find the mass of Ag formed:

0.0353 mols AgNO₃ x 2 mols Ag / 2 mol AgNO₃ x 108 g Ag / mol Ag = 3.81 g Ag formed

Answer 2

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Step 1. Chemical Equation

Zn_(s) + 2AgNO_3(aq) ---> Zn(NO₃)_2(aq) + 2Ag_(s)

Step 2. Find Limiting reagent

Moles of Ag from Zn = 2.00 g of Zn /(65.39 g Zn/mol Zn)* (2 mol of Ag/ 1 mol of Zn) = 0.0611₇₁ mol of Ag

Moles of Ag from AgNO₃= 6.00 g of AgNO3 /(169.87 g AgNO₃/mol AgNO₃)* (2 mol of Ag/ 2 mol of AgNO₃) = 0.0353₂₁ mol of Ag

Step 3. Find the mass

Mass of Ag = 0.0353₂₁ mol of Ag* (107.9 g/mol) = 3.81 g of Ag

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