Here we will use the general formula of Nernst equation:
Ecell = E°Cell - [(RT/nF)] *㏑Q
when E cell is cell potential at non - standard state conditions
E°Cell is standard state cell potential = - 0.87 V
and R is a constant = 8.314 J/mol K
and T is the temperature in Kelvin = 73 + 273 = 346 K
and F is Faraday's constant = 96485 C/mole
and n is the number of moles of electron transferred in the reaction=2
and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)
so by substitution :
0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q
∴ Q = 4.5 x 10^-26
Answer:
4.5 moles of lithium sulfate are produced.
Explanation:
Given data:
Number of moles of lead sulfate = 2.25 mol
Number of moles of lithium nitrate = 9.62 mol
Number of moles of lithium sulfate = ?
Solution:
Chemical equation:
Pb(SO₄)₂ + 4LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄
Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.
Pb(SO₄)₂ : Li₂SO₄
1 : 2
2.25 : 2/1×2.25 = 4.5 mol
LiNO₃ : Li₂SO₄
4 : 2
9.62 : 2/4×9.62 = 4.81 mol
Pb(SO₄)₂ produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of Li₂SO₄.
The correct answer is A, B and C
The correct answer you are looking for is napalm
Hope this helps!
The reaction will be: FeBr2 + K --> KBr + Fe
Balancing gives: FeBr2 + 2K --> 2KBr + Fe
The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol.
We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2
Based on stoichiometry:
(0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr
