The balanced equation
for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃OH(g)
The given
concentrations are at equilibrium state. Hence we can use them directly in
calculation with the expression for the equilibrium constant, k.
expression for k can be written as
k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²
</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M
From substitution,
k = 0.030
M / 0.020 M x (0.072 M)²
k =
289.35 M⁻²
<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².
<span> </span>
Answer:
Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)
Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles
Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles
Limiting reagent is Na₂SO₄ as it controls product formation
Moles of HgSO₄ formed = 0.1172 moles
= 0.1172 x 296.65
= 34.757g
Explanation:
When the neutrons and electrons are the same. For example, sodium (Na) has an atomic mass of 11, meaning it has 11 protons and 11 electrons etc.
