Mole ConceptHard
Question
A quantity of 27.6 g of K2CO3 was treated by a series of reagent so as to convert all of its carbon to K2Zn3[Fe(CN)6]2. The mass of the product formed is (K = 39, Zn = 65.4, Fe = 56)
Options
A.139.2 g
B.11.6 g
C.69.6 g
D.23.2 g
Solution
$12K_{2}CO_{3} \rightarrow K_{2}Zn_{3}\left\lbrack Fe(CN)_{6} \right\rbrack_{2}$
12 × 138 g 698.2 g
$\therefore 27.6g \frac{698.2}{12 \times 138} \times 27.6 = 11.64g$
Create a free account to view solution
View Solution FreeMore Mole Concept Questions
The solubility of Mg(OH)2 is S moles/litre. The solubility product under the same condition is...Law of multiple proportions is not applicable for the oxide(s) of...A 2 m long tube closed at one end is lowered vertically into water until the closed end is flushed with the water surfac...2SO2 + O2 ⇋ 2SO3 initially 4 moles each of SO2 & O2 are present, at equilibrium 25% of O2 is used. Total mols at e...An amount of 1.0 × 10–3 moles of Ag+ and 1.0 × 10–3 moles of CrO4 2− reacts together to form solid Ag2CrO4. What is the ...