Question

How many moles of Silver (1) hydrogen carbonate are produced when 167 g of Tin (Il) hydrogen carbonatecombines with an excess amount of silver (I) dichromate? Use the following balanced equation:1 Sn(HCO3)2 + 1 AgzCr20, ---> 1 SnCr207 + 2 AgHCO3

Answer 1

Step 1

The reaction:

1 Sn(HCO3)2 + 1 Ag2Cr2O7 => 1 SnCr207 + 2 AgHCO3 (completed and balanced)

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Step 2

Information provided:

167 g of Tin (Il) hydrogen carbonate, Sn(HCO3)2 => the limiting reactant

The excess = Ag2Cr2O7

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Information needed:

The molar mass of Sn(HCO3)2 = 240.7 g/mol (use your periodic table please)

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Step 3

By stoichiometry,

1 mol Sn(HCO3)2 = 240.7 g

Procedure:

1 Sn(HCO3)2 + 1 Ag2Cr2O7 => 1 SnCr207 + 2 AgHCO3

240.7 g Sn(HCO3)2 ---------- 2 x 1 mole AgHCO3

167 g Sn(HCO3)2 ---------- X

X = 167 g Sn(HCO3)2 x 2 x 1 mole AgHCO3/240.7 g Sn(HCO3)2

X = 1.39 moles

Answer: 1.39 moles AgHCO3 are produced