[tex]NH_3+HNO_3-> NH_4NO_3[/tex]
1 mole of nitric acid produce 1 mole of ammonium nitrate.
moles in 5000 kg of ammonium nitrate :
[tex]n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles[/tex] ( molecular mass of ammonium nitrate is 80 gm/mol )
So, number of moles of nitric acid required are also 62500 moles.
Mass of 62500 moles of nitric acid :
[tex]mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg[/tex]
Hence, this is the required solution.
