The powder mixture of magnesium and silver weight 1,5 g treated with excess of a solution of sulfuric acid ,wherein stood 560 ml (normal conditions)

m(mixture) =1,5 g
V (Н2) = 560 ml
find:
ω(Мg) -?
ω(Аg) -?
Solution:
Мg +  Н2SО4  = МgSО4  +  Н2↑
Silver does not react :
  Аg +  Н2SО4(dil.)   ↛
ν(Н₂)=(V(Н₂))/V_m = (0,56 л)/(22,4 л/моль)=0,025 mol
ν(Н2) = ν(Мg) ⇒ν(Мg)  = 0,025 mol
М(Мg) = 24 g/mol
m(Мg) = ν(Мg) • М(Мg) = 0,025 mol • 24 g/mol  = 0,6 g
m(Аg) = m(mixture) - m(Мg)  = 1,5 g – 0,6 g = 0,9 g
ω(Мg)=(m(Мg))/(m(mixture))= (0,6 g)/(1,5 g)=0,4,или 40%
ω(Аg)=(m(Аg))/(m(mixture))= (0,6 g)/(1,5 g)=0,4,или 40%
Answer:ω(Мg) = 40 %; ω(Аg) = 40 %.