Question

Calculate the mass of aluminum sulfide produced when 15.60 mL of 0.2250 M potassium sulfide is mixed with excess aluminum nitrate according to the given balanced equation: 2 Al(NO3)3(aq) + 3 K2S(aq) → 6 KNO3(sq) + Al2S3(s)

Answer 1

moles of K2S reacting = Molarity of K2S * volume of K2S in L

= 0.2250 M * 0.01560 mL

= 0.00351 mol

from reaction,

mol of Al2S3 produced = (1/3)*moles of K2S reacting

= (1/3)*0.00351 mol

= 0.00117 mol

Molar mass of Al2S3 = 2*MM(Al) + 3*MM(S)

= 2*26.98 + 3*32.07

= 150.17 g/mol

we have below equation to be used:

mass of Al2S3,

m = number of mol * molar mass

= 1.17*10^-3 mol * 150.17 g/mol

= 0.176 g

Answer: 0.176 g