From the equation, we see that the molar ratio of Fe : S required is:
8 : 1
The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265
The molar ratio is:
1 : 1.6
Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of
Fe : FeS is
1 : 1
So 0.168 moles of FeS will form. The mass of FeS will be:
Mass = 0.168 * (56 + 32)
Mass = 14.78 grams
14.78 grams of FeS will be formed.
C = Q / M * ΔT
Δf - Δi = 42.0ºC - 12.0ºC = 30.0ºC
C = 226 J / 58.3 * 30.0
C = 226 / 1749
C = 0.129 J/gºC
hope this helps!
<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
