Question

How many grams of Silver (1) hydrogen carbonate are produced when 63.5 g of Tin (I) hydrogen carbonatecombines with 47.3 g of Silver (I) dichromate? Use the following balanced equation:1 Sn(HCO3)2 + 1 AgCr2O -> 1 SnCr2O7 + 2 AgHCO3

Answer 1

Step-by-step explanation:

First, let's rewrite the equation:

1 Sn(HCO3)2 + 1 AgCr2O -> 1 SnCr2O7 + 2 AgHCO3

Now, let's transform grams into moles of each compound using the folloing formula:

n = m/MM

MM of Sn(HCO3)2 = 240.7437

MM of AgCr2O = 227.8598

n of Sn(HCO3)2 = 63.5/240.7437 = 0.264 moles

n of of AgCr2O = 47.3/ 227.8598 = 0.208 moles

AgCr2O is the limiting reactant, so:

1 mole AgCr2O ---- 2 moles AgHCO3 (reaction equation gives us this information)

0.208 moles ---- x moles

x = 0.415 moles of AgHCO3

Now let's transform to gram:

m = n*MM

m = 0.415*168.8850

m = 70.1 g

Answer: m = 70.1 g