Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be
FRANCIUM, which is at the base of Group One.
Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.
I think the amount would be a 0.4998 mol
I did moles=mass(g)/A,r
=12.5/24.3 to get that
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:
Is it prescribe to you?If so than yes if not then no need to
Explanation:
Answer:
D
Explanation:
I think but it is an better attempt than the other guy answer.
