Balance the equation first:
2 Fe+6 HNO3→2 Fe(NO3)3+3H2
Then calculate mass of Iron :
4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe
Now use Stoichiometry:
371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3
Convert moles of nitric acid to grams
19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3
It seems like the answer would be supporting services.