The % yield of Hf is 91. 8%
calculation
% yield = actual mass/ theoretical mass x100
actual mass = 2.85 Kg
calculate the theoretical mass
by first write the equation for reaction
CaF2 +H2SO4 → 2HF + CaSO4
find the moles of CaF2 used
moles = mass/molar mass
mass = 6.05Kg = 6.05 x1000 =6050 Kg
molar mass of CaF2 = 40 + (19 x2) = 78 g/mol
moles= 6050/ 78 = 77.6 moles
by use of mole ratio between CaF2 to HF which is 1:2 the moles of HF is therefore = 77.6 x2 =155.2 moles of Hf
find the theoretical mass of HF = moles x molar mass ( 1 +19=20g/mol)
= 155.2 moles x 20 g/mol = 3104 grams = 3104 /1000 = 3.104 Kg
The % yield is therefore = 2.85 Kg/ 3.104 Kg x100 = 91.8%
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
Answer:
it'a answer number 2
Explanation: you divide the mass by volume and 32.2 divided by 4 is 8.05
Moles=concentration x volume. 1.5/1000 multiply 0.450 = 6.75x10^-4
To find the mass of fecl3, we have the formula, mass=moles x molar mass. 6.75x10^-4 x (55.8+35.5 x 3). = 0.1096g
Not sure tho if i did it right
