Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
5.15 moles
Explanation:
2zn + o2 = 2zno
5.15 2.57 5.15 moles
nzno=500/(16x2+65)= 5.15 moles
-> nzn = 5.15 x 2 ÷ 2 = 5.15 moles
i think it is 6 valence electrons
It showed that the nucleus is positively charged, and that the atom isn't just made of electrons.
2 SO₃ --> 2 SO₂ + O₂
I 12 0 0
C -2x +2x +x
---------------------------------------------
E 12-2x 2x x
Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5
The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]²
where the unit used is conc in mol/L.
K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
<em>K = 0.0556</em>
