Question

Nitric acid and zinc react to form zinc nitrate, ammonium nitrate, and water. 4 Zn(s) + 10 HNO3(aq) - 4 Zn(NO3)2(aq) + NH4NO3(aq)+ 3 H20(0) (a) How many atoms of zinc react with 1.05 g HNO3? atoms (b) Calculate the number of grams of zinc that must react with an excess of HNO3 to form 28.3g NH4NO3

Answer 1

   4 Zn(s) + 10 HNO3(aq) ---------------- 4 Zn(NO3)2(aq) + NH4NO3(aq) + H2O(l)

4 mole    10 mole 1 mole

mass of HNO3 = 1.05 grams

molar mass of HNO3 = 63.0 gram/mole

number of moles of HNO3 = mass/molar mass = 1.05/63.0 = 0.0167 moles

one mole of an atom contains Avagadro number of atoms

Avagadro's number = 6.023x10^23

according to equation

10 moles of HNO3 = 4 moles of Zn

0.0167 moles of HNO3 =?

                                     = 4x0.0167/10 = 0.00668 moles of Zn

number of moles of Zn reacts = 0.00668 moles

number of Zn atom present in 1 mole of Zn = 6.023x10^23 atoms

number of Zn atom present in 0.00668 moles of Zn = ?

                                                                                = 0.00668 x 6.023x10^23= 0.040x10^23 atom

Number of Zn atoms = 4.0x10^21 atoms

b)

according to the equation

1 mole of NH4NO3 = 4 moles of Zn

molar mass of NH4NO3 = 80.04 gram/mole

molar mass of Zn= 65.38 gram/mole

according to equation

80.04 grams of NH4NO3 = 4x65.38 grams of Zn

28.3 grams of NH4NO3 = ?

                                           = 4x65.38 x 28.3/80.04 = 92.47 grams

Mass of Zn = 92.47 grams.