Question

If 501 grams of Ni(OH2)62+ reacts with 108 grams of ammonia, two products are formed. One product is water and the other one is an unknown. How many grams of the unknown compound will be formed? Assume 100% yield. (Hint: you do not need to know the formula for the unknown compound) SHOW STEPS PLEASE

Ni(OH2)62+(aq) + 5NH3(aq) = Unknown (aq) + 5H2O(l)

Answer 1

Ni(OH2)62+(aq) + 5NH3(aq) = Unknown (aq) + 5H2O(l)

according to balanced reaction

333.57 g Ni(OH2)62+ reacts with 5 x 17 g NH3

501 g Ni(OH2)62+ reacts with 501 x 5 x 17 / 333.57 = 127.7 g NH3

but we have only 108 g NH3 so NH3 is limiting reagent

5 x 17 g NH3 reacts with 333.57 g Ni(OH2)62+

108 g NH3 reacts with 108 x 333.57 / 5 x 17 = 36025.56 / 85 = 423.83 g Ni(OH2)62+

total mass of reactants reacted = 108 + 423.83 = 531.83 g

so 531.83 g products must be produced

5 x 17 g NH3 produce 5 x 18 g H2O

108 g NH3 produce 108 x 5 x 18 / 5 x 17 = 9720 / 85 = 114.35 g H2O

mass of unknown = total mass of products - mass of water = 531.83 - 114.35 = 417.48 g

mass of unknown = 417.48 g