Question

Potassium sulfide reacts with cobalt(II) nitrate in an aqueous solution. Write a balanced equation for this double displacement reaction. Calculate the mass of the solid product produced from 175 mL of a 0.225 M potassium sulfide combined with 250. mL of 0.180 M cobalt (II) nitrate.

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Answer 1

4.29 g

Step-by-step explanation:

The equation of the reaction is; Co(NO3)2(aq) + K2S(aq) → CoS(s) + 2KNO3(aq)

Number of moles of potassium sulphide = 175/1000 L * 0.225 M = 0.039 moles

Number of moles of cobalt (II) nitrate =250/1000 L * 0.180 M = 0.045 moles

Hence; potassium sulphide is the limiting reactant

Since the mole ratio is 1:1

Mass of solid product = 0.039 moles * 110 g/mol = 4.29 g