1) Chemical equation
2Al + 6 HCl ---> 2Al Cl3 + 3 H2
2) molar ratios
2 mol Al : 3 moles H2
3) Proportion
2 mol Al / 3mol H2 = x / 9 mol H2
4) Solve for x
x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag
Answer: 6 moles
Answer:
Three product with are SO2, H2O and CuSO4
Explanation:
Answer:
[Cl⁻] = 0.016M
Explanation:
First of all, we determine the reaction:
Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)
This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:
PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps
Initial x
React s
Eq x - s s 2s
As this is an equilibrium, the Kps works as the constant (Solubility product):
Kps = s . (2s)²
Kps = 4s³ = 1.7ₓ10⁻⁵
4s³ = 1.7ₓ10⁻⁵
s = ∛(1.7ₓ10⁻⁵ . 1/4)
s = 0.016 M
I believe the correct response is A. A solution with a pH close to 14 is considered a strong base.
