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!
Answer:
55.85 grams of Fe is formed.
Explanation:
Identify the reaction:
2Fe₂O₃ + 3C → 4Fe + 3CO₂
Identify the limiting reactant, previously determine the mol of each reactant
(mass / molar mass)
10 g / 12 g/m = 0.83 moles C
80 g / 159.7 g /m = 0.500 moles Fe₂O₃
2 moles of oxide need 3 moles of C, to react
0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol
I have 0.83 moles of C, so C is the excess.
The limiting is the oxide.
3 mol of C need 2 mol of oxide to react
0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)
Ratio is 2:4 (double)
If I have 0.5 moles of oxide, I will produce the double, in the reaction.
1 mol of Fe, will be produce so its mass is 55.85 g
the answer is magnetic separation, not sedimentation separation
28%
Explanation:
mass of solute(KBr) = 3.73g
mass of solvent(H2O) = 131g
mass of solution = mass of solute + mass of solvent
= 3.73 + 131
= 134.73g
