If 34.7 g of agno₃ react with 28.6 g of h₂so₄ according to this unbalanced equation below, how many grams of ag₂so₄ could be formed? agno₃(aq) + h₂so₄ (aq) → ag₂so₄ (s) + hno₃ (aq)

The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams

    calculation

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)

Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

   the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles
The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

     AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles
    finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

Ithink d hope that