Question

For following reaction, calculate the grams of indicated product when 16.9 g of the first reactant and 10.4 g of the second reactant is used:
Fe2O3(s)+3H2(g)→2Fe(s)+3H2O(l) (H2O)

m(H2O) = ???? Thank you!

Answer 1

n(Fe₂O₃) = m/M = 16.9 / (56*2+16*3) = 0.1056 (mol)
n(H₂) = 10.4/2 = 5.2 (mol) – excess
m(H₂O) = n*M = 3*n(Fe₂O₃) * M(H₂O) = 3*0.1056*(16+1*2) = 5.7024 (g)

Answer
m(H₂O) = 5.7024 g