How do solve this

Question

Mercury(II) nitrate is a compound formerly used to treat fur to make felt for hats. It can also cause a psychological illness, which is where the phrase 'mad as a hatter' originated. Mercury(II) nitrate is synthesized by the reaction of mercury metal with hot concentrated nitric acid: Hg(l) + 2 HNO₃(aq) → Hg(NO₃)₂ + H₂(g) What mass in grams of mercury metal would be required to form 6.55 grams of mercury(II) nitrate?

Answer 1

Start with a balance chemical reaction equation. In this case, you were given that equation:

Hg(l) + 2HNO₃(aq) → Hg(NO₃)₂ + H₂(g)

This equation is in "units of moles", i.e., 1 mole of Hg plus 2 moles of HNO₃ produces 1 mole of Hg(NO₃)₂ plus 1 mole of H₂

Since you have been given the mass of the product (grams) and not the number of moles, this chemical reaction equation is not useful to you until you change the mass into the number of moles. You need the molar mass to do that. To calculate the molar mass of Hg(NO₃)₂ you must look up all its components on the periodic table:

Hg: 200.59 x 1 atom = 200.59 g/mol

N: 14.01 x 2 atoms = 28.02 g/mol

O: = 15.999 x 6 atoms = 95.99 g/mol

Hg(NO₃)₂ = 200.59 + 28.02 + 95.99 = 324.6 g/mol

To determine the number of moles: (6.55 g)/(324.6 g/mol) = 0.0202 moles

Looking at the chemical reaction equation, it takes 1 mole of Hg to make 1 mole of Hg(NO₃)₂ so 0.0202 moles of Hg will produce 0.0202 moles of Hg(NO₃)₂

So we know that we need 0,0202 moles of Hg. To find the mass of this, use the atomic mass of mercury:

(0.0202 moles)(200.59 g/mol) = 4.05 grams of Hg

So 4.05 grams of Hg is required to produce 6.55 grams of Hg(NO₃)₂