Question

what is the theortical yield of beryllium needed to produce 36.0g of hydrogen ? (assume an excess of water)
Be(s)+2H2O(L)->Be(OH)2(aq)+H2(g)

Answer 1

n(Be)/n(H2) = 1/1
n(Be)=n(H2)
n(Be) = 36 g / 2,016 gmol-1 = 17,85 mol
m(Be) = 17,85 mol * 9,01 gmol-1 = 160,83 g