H₂SO₄:
V=0,95L
Cm=0,420mol/L
n = CmV = 0,42mol/L * 0,95L = 0,399mol
KOH:
V=0,9L
Cm=0,26mol/L
n = CmV = 0,26mol/L * 0,9L = 0,234mol
H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O
1mol : 2mol
0,399mol : 0,234mol
limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol
n = 0,282mol
V = 0,950L + 0,900L = 1,85L
Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
Seems right, but if it’s just asking for one i would pick B
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
scandium is a transition metal and shows a valency of 3.
