Answer to Question #98784 in Chemistry for Lesly Coss

Question

Consider the reaction K2S(aq) + Co(NO3)2(aq) --> 2KNO3(aq) + CoS(s)

What mass of CoS is produced from 4.84 g Co(NO3)2 and 115 mL of 0.225 M K2S solution? _____ g (Give answer to correct sig figs in units of g)

Answer 1

Answer on Question #98784 – Chemistry – Other

Task:

Consider the reaction K₂S(aq) + Co(NO₃)₂(aq) --> 2KNO₃(aq) + CoS(s)

What mass of CoS is produced from 4.84 g Co(NO₃)₂ and 115 mL of 0.225 M K₂S solution? _____ g.

(Give answer to correct sig figs in units of g).

Solution:

M(K₂S) = 2*Ar(K) + Ar(S) = 2*39.098 + 32.06 = 110.256 g/mol;

M(Co(NO₃)₂) = Ar(Co) + 2*Ar(N) + 6*Ar(O) = 58.933 + 2*14.007 + 6*15.999 = 182.941 g/mol;

M(CoS) = Ar(Co) + Ar(S) = 58.933 + 32.06 = 90.993 g/mol.

n(Co(NO₃)₂) = m(Co(NO₃)₂) / M(Co(NO₃)₂) = 4.84 / 182.941 = 0.026456 mol;

n(K₂S) = C(K₂S) * V(K₂S) = 0.225 * 0.115 = 0.025875 mol;

According to the reaction equation:

Each of these reacts in a 1:1 mole ratio to form the product CoS, so if we use up all of the reactant, we should be able to produce that many moles of CoS. However, in our case, K2S - limiting reactant [n(K₂S) < n(Co(NO₃)₂)].

So we should only be able to make 0.025875 mol of CoS product.

Then,

n(K₂S) = n(CoS);

n(K₂S) = m(CoS) / M(CoS);

m(CoS) = M(CoS) * n(K₂S) = 90.993 * 0.025875 = 2.3544 g

m(CoS) = 2.35 g.

Answer: m(CoS) = 2.35 g.