The percentage of yield was 777.78%
<u>Explanation:</u>
We have the equation,
Be
[s] + 2
HCl
[aq] → BeCl
2(aq] +
H
2(g] ↑ Be
(s] +
2
HCl
[aq] → BeCl
2(aq] +
H
2(g]
↑
To find the percent yield we have the formula
Percentage of Yield= what you actually get/ what you should theoretically get x 100
=3.5 g/0.45 g 100
= 777.78 %
The percentage of yield was 777.78%
A catalyst speeds up the rate of reaction so the answer is B.
The molarity and normality of 5.7 g of Ca(OH)2 in 450ml 0f solution is calculated as follows
molarity = moles/volume in liters
moles =mass/molar mass
= 5.7g/74g/mol = 0.077moles
molarity = 0.077/450 x1000= 0.17M
Normality = equivalent point x molarity
equivalent point of Ca(OH)2 is 2 since it has two Hydrogen atom
normality is therefore = 0.17 x2 = 0.34 N
