Question

Question 1:

Be sure to answer all parts. Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 622 g of iodine trichloride reacts with 124.1 g of water:

ICl3 + H2O → ICl + HIO3 + HCl [unbalanced]

mol HIO3?

g HIO3?

What mass of the excess reactant remains in grams?

Question 2: Be sure to answer all parts. When 0.107 mol of carbon is burned in a closed vessel with 5.06 g of oxygen, how many grams of carbon dioxide can form? g CO2 ? Which reactant is in excess, and how many grams of it remain after the reaction? g (select carbon or oxygen)?

Answer 1

1)

2 ICl3    +    3 H2O -----------> ICl +    HIO3 + 5 HCl

466.5 g       54.06 g                                175.9 g

622 g            124.1 g                                  ??

466.5 g ICl3   -----------> 54.06 g H2O

622 g ICl3    ------------> ??

mass of H2O needed = 622 x 54.06 / 466.5 = 72.08 g

but we have 124.1 g . so this is excess .

limiting reagent is ICl3.

466.5 g ICl3   -----------> 175.9 HIO3

622 g ICl3    ------------> ??

mass of iodic acid formed = 622 x 175.9 / 466.5

mass of iodic acid formed = 235 g

moles of iodic acid formed = 235 / 175.9 = 1.33 mol

excess reactant remain = 52.02 g

2)

C (s)     +   O2 (g)    ---------------->   CO2 (g)

12 g           32 g                               44.01 g

1.284 g      5.06 g                              ??

mass of CO2 formed = 4.71 g

O2 is excess.

mass of remaining amount = 1.64 g