Question

How many moles of zine nitrate are formed from 52.0 grams of zinc and unlimited nitric acid (Hint: What does "unlimited tell you 6. When 1.25 g of mercury (II)oxide is heated, what is the theoretical yield of mercury? Suppose 1.09 g of mercury is actually collected. What is the percent yield? 2Hgo. → 2 Hg. + 0.

Answer 1

1. Balanced equation for this reaction is ,

Zn + 2HNO3 ------> Zn(NO3)2 + H2

1 mole of Zn gives 1 mole of zinc nitrate.

Number of moles are given as,

Number of moles = Mass/Molar mass

Molar mass of Zn = 65.4 g/mol

Moles of Zn = 52/65.4

= 0.80 moles of Zn

Moles of Zinc nitrate formed = 0.80 moles

2. 2 moles of HgO gives 2 moles Hg

1 mole of HgO will give 1 mole of Hg

Molar mass of HgO = 216.59 g/mol

Moles of HgO = 1.25/216.59

= 0.0058 moles

Moles of Hg formed = 0.0058 mol

Molar mass of Hg = 200.59 g/mol

Mass of Hg formed = 200.59 g/mol × 0.0058 mol

Theoretical yield = 1.16 grams of Hg

Percent yield = ( actual yield/theoretical yield) × 100

Percent yield = (1.09/1.16) × 100

= 93.97 %