Maximum ! if copper metal pieces were added to an aqueous solution of silver nitrate, the silver would be replaced in a single replacement reaction forming aqueous copper (ii) nitrate and solid silver. 1. write a balanced chemical equation for the reaction.2. how much silver is produced is 15.00 grams of cu is added to the solution of excess silver nitrate? 3. if silver metal sells for $4.50/ounce, how much would the silver collected be worth? (1 gram = 0.0353 oz)

  1.The balanced  chemical equation is as below

Cu +  2 Ag(NO)₃ → Cu(NO₃)₂  +2 Ag

2.The  much silver  produced  is  50.9 g

Step 1: find the  moles of Cu

moles =  mass÷  molar mass

  from periodic table the molar mass of Cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236  moles

Step 2: use the mole ratio to determine the  moles of Ag

from equation above   Cu:Ag  is 1:2

therefore the moles of Ag   =0.236  moles  x 2/1 =0.472  moles

Step 3: find the mass  of Ag

mass = moles x molar mass

from periodic table  the molar mass of Ag =107.87 g/mol

mass  = 0.472  moles x 107.87 g/mol = 50.9 g

3.  The  much  silver  collected worth is  1.797 oz

 convert  50.9 g  to oz

that is  1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1  gram}  =1.797 oz

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2