Hey there!:
Molar mass of Mg(OH)2 = 58.33 g/mol
number of moles Mg(OH)2 :
moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles
Molar mass of H3PO4 = 97.99 g/mol
number of moles H3PO4:
moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles
Balanced chemical equation is:
3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O
3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !
Now , we will use Mg(OH)2 in further calculation .
Molar mass of Mg3(PO4)2 = 262.87 g/mol
According to balanced equation :
mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2
= (1/3)*0.5246
= 0.1749 moles of Mg3(PO4)2
use :
mass of Mg3(PO4)2 = number of mol * molar mass
= 0.1749 * 262.87
= 46 g of Mg3(PO4)2
Therefore:
% yield = actual mass * 100 / theoretical mass
% = 34.7 * 100 / 46
% = 3470 / 46
= 75.5%
Hope that helps!
Ionic is the answer. This is because lithium has a positive charge, while chlorine has a negative charge, meaning the compound doesn’t necessarily have an overall charge.
Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
<h3>Why does it take this much energy to boil the water?</h3>
We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.
Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
To learn more about the behavior of water visit:
brainly.com/question/1416592?referrer=searchResults
The C stands for Specific Heat
