Pure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in jewelry, platinum is combined with other metals to form an alloy. To determine the amount of platinum in an alloy, a 8.528 g sample of an alloy containing platinum and cobalt is reacted with excess nitric acid to form 2.49 g of cobalt(II) nitrate. Calculate the mass percent of platinum in the alloy.

percentage mass of platinum in the alloy ≈ 90.60 %

Explanation:

The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.

The chemical reaction can be represented as follows:

Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) +  NO₂ (l) + H₂O (l)

The balanced equation

Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)

Cobalt is the limiting reactant

atomic mass of cobalt = 58.933 g/mol

Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933  g

58.933 g of cobalt produce 182.933  g of  Co(NO₃)₂

? gram of cobalt will produce 2.49 g of Co(NO₃)₂

cross multiply

grams of cobalt that will react = (58.933 × 2.49)/182.933

grams of cobalt that will react = 146.74317000/182.933

grams of cobalt that will react= 0.8021689362 g

grams of cobalt that will react = 0.802 g

mass of platinum in the alloy  = 8.528 g - 0.802 g = 7.726 g

percentage  mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %

percentage mass of platinum in the alloy ≈ 90.60 %

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.