Molar mass
H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol
Identifying excess reagent and the limiting of the reaction :
8 H2S(g) + 4 O2(g) = S8(I) + 8 H2O(g)
8 x 34 g H2S --------> 256. 52 g S8
35.0 g ----------------> ??
35.0 x 256.52 / 8 x 34 =
8978.2 / 272 => 33.00 g of S8
H2S is the limiting reactant
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4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??
40.0 x 256.52 / 4 x 31.99 =
10260.8 / 127.96 = 80.16 g of S8
O2 is the excess reagent is the excess <span>reagent
</span>
------------------------------------------------------------
H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced
33.0 g ----------- 100%
?? g ------------- 95 %
95 x 33.00 / 100 => 31.35 g
hope this helps!
Answer:
- <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>
Explanation:
Assume 100% ionization:
<em><u>1) 0.100 M solution K₂SO₄</u></em>
- K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)
- Mole ratios: 1 mol K₂SO4 : 2 mol K⁺ + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3
- At constant volume, the mole ratios are equal to the concentration ratios (M).
- 1 M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions
This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.
<u>2) 0.0800 M Na₂CO₃</u>
- Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻
- 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions
This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.
<u>3) 0.100 M NaCl </u>
- 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions
This is not equal to 0.300 M ion, so not a correct option.
<u>4) 0.0750 M Na₃PO₄</u>
- 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions
Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.
<u>5) 0.0500 M NaOH </u>
- 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions
Not equal to 0.300 M, so wrong choice.
Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of 
will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since is also a reactant, its concentration will decrease.
Answer:
Option : KBr + I -> KBr+I
Explanation:
Single-replacement reaction or single displacement reactions are a type of chemical reactions in which a whole compound reacts with an element in such a way that the element takes place of one of the compound's own elements and sets it free.
If we talk about KBr and I displacement reaction is not possible among these because Iodine is less reactive than Bromine that is why it will not react with KBr or replace Br.
KBr + I -> KBr+ I
Potassium Bromide + Iodine -> Potassium bromide + Iodine
Hope it help!
2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2
2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺
2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻
