A piece of copper of mass 300g at a temperature of 950 o C is quickly transferred into a vessel of negligible heat capacity containing 250g
of water at 25 o C. If the final steady temperature of the mixture is
100 o C. Calculate the mass of water that will boil away. (take specific
heat capacity of copper = 4.0x10 2 Jkg -1 K -1 , specific heat capacity of
water = 4.2 x10 3 Jkg -1 K -1 , specific latent heat of vaporization of steam =
2.26x10 6 Jkg -

  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