Question

For each of the following reactions, calculate the grams of indicated product when 25.5 g of the first reactant and 40.0 g of the second reactant is used:
3Fe(s)+4H2O(l)→Fe3O4(s)+4H2(g) (Fe3O4)

Answer 1

m(Fe) = 25.5 g

m(H2O) = 40 g

n(Fe) = m/M = 25.5 g / 56 g/mol = 0.455 mol

n(H2O) = 40 g / 18 g/mol = 2.222 mol

Fe is the limiting reactant.

n(Fe3O4) = n(Fe)/3 = 0.455/3 = 0.152 mol

M(Fe3O4) = 56*3 + 16*4 = 232 g/mol

m(Fe3O4) = n*M = 0.152mol*232g/mol = 35.264 g