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>
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
Answer:
63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.
Explanation:
I this is college level chemistry not regular high school chem.
