Answer:
Mass = 18.9 g
Explanation:
Given data:
Mass of Al₂O₃ formed = ?
Mass of Al = 10.0 g
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 27 g/mol
Number of moles = 0.37 mol
Now we will compare the moles of Al and Al₂O₃.
Al : Al₂O₃
4 : 2
0.37 : 2/4×0.37 = 0.185 mol
Mass of Al₂O₃:
Mass = number of moles × molar mass
Mass = 0.185 mol × 101.9 g/mol
Mass = 18.9 g
Answer:
V₂ = 1.86 L
Explanation:
Given data:
Initial volume = 4.30 L
Initial pressure = 1 atm
Initial temperature = 273.15 K
Final temperature = 302 K
Final volume = ?
Final pressure = 2.56 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂
/T₁ P₂
V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm
V₂ = 1298.6 atm.L.K / 699.26 K.atm
V₂ = 1.86 L
M= moles de soluto / litros de solucion
moles de soluto = M. litros de solucion
Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3