Suppose 100 g of PbS is reacted with 4g of C and 20g O2.
Find the limiting reactant for this reaction.
2PbS(s) + 2C(s) + 3O2(g) → 2Pb(s) +2CO(g) + 2SO2(g)
n(PbS) = 100 g / (2 *239.3 g/mol) = 0.21 mol
n(C) = 4 g / (2*12 g/mol) = 0.17 mol
n(O2) = 20 g / (3*32 g/mol) = 0.21 mol
C is limiting reactant
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