Product:
C: 1
O:3
H: 2
reactant:
C: 8
H: 18
O: 2
first you need to put probably an eight in front of the Carbon.
Next, add a coefficient of 9 to Hydrogen
finally, add a 3 INSTEAD of a 2 to Oxygen
should look like:
C8H18+O3>C8O2+H9O
Answer:
The concentration of the unknown acid (HA) is 0.434M
The molar mass of HA is 13.3g/mole
Explanation:
DETERMINATION OF MOLARITY OF THE UNKNOWN ACID
CaVa/CbVb = Na/Nb
From the equation of reaction and at equivalence point, Na = Nb = 1
Therefore, CaVa = CbVb
Va (volume of acid solution) = 20mL = 20/1000 = 0.2L
Cb (concentration of KOH) = 0.715M
Vb (volume of KOH) = 12.15mL
Ca (concentration of acid) = CbVb/Va
Ca = 0.715M × 12.15mL/20mL = 0.434M
DETERMINATION OF MOLAR MASS OF HA
Number of moles of acid = concentration of acid × volume of acid solution in liters = 0.434 × 0.2 = 0.0868mole
Molar mass of HA = mass/number of moles = 1.153g/0.0868mole = 13.3g/mole
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Actual yield of Fe2(So4)3 = 18.5g
2FePo4 + 3Na2SO4 -> Fe2(SO4)3 + 2Na3PO4
Mole of FePO4 = mass of it / its molar mass =
25 g / (55.8 + 31 + 16*4) = 0.166 mol
every 2 mole of FePO4 will form 1 mole of Fe2(SO4)3
Mole of Fe2(SO4)3 produced = 0.166 / 2 = 0.0829 mol
0.0829 * (55.8*2 + 3*(32.1+ 16*4)) = 33.148 g of Fe2(SO4)3
18.5 / 33.148 * 100 = 55.8%
