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:
i think i like 14 or 8 atom
Explanation:
Methane is a hydrocarbon which when burns in air (combustion) produces carbon dioxide and water. The equation for the reaction;
CH4 +2O2 = CO2 +2H2O
When one mole of methane combusts 2 moles of water are formed
Therefore; when 22 moles of methane combusts 44 moles of water are formed (22 ×2)
Answer:
[NaCH₃COO] = 2.26M
Explanation:
17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)
Let's determine the volume of solution, by density
Mass of solution / Volume of solution = Solution density
100 g / Volume of solution = 1.09 g/mL
100 g / 1.09 g/mL = 91.7 mL
17 grams of solute is contained in 91.7 mL
Molarity (M) = Mol of solute /L of solution
91.7 mL / 1000 = 0.0917L
17 g / 82 g/m = 0.207 moles
Molariy = 0.207 moles / 0.0917L → 2.26M
Answer:
The statement describes a process involved in the evolution of Earth’s early atmosphere would be:
Cyanobacteria transformed carbon dioxide in the atmosphere into oxygen during photosynthesis
Hope it helped :3
