Answer:
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The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.
0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
Answer:
pH = 10.75
Explanation:
To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]
Using the equation:
pH = 14 - pOH
We can find the pH of the solution.
The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M
pOH is
pOH = -log 5.6x10⁻⁴M
pOH = 3.25
pH = 14-pOH
<h3>pH = 10.75</h3>
Answer:
Option A. FeCl3
Explanation:
The following data were obtained from the question:
Mass of iron (Fe) = 6.25g
Mass of the compound formed = 18g
From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow
Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.
Mass of chlorine (Cl) = 18 – 6.25
Mass of chlorine (Cl) = 11.75g
The compound therefore contains:
Iron (Fe) = 6.25g
Chlorine (Cl) = 11.75g
The empirical formula for the compound can be obtained by doing the following:
Step 1:
Divide by their molar mass
Fe = 6.25/56 = 0.112
Cl = 11.75/35.5 = 0.331
Step 2:
Divide by the smallest
Fe = 0.112/0.112 = 1
Cl = 0.331/0.112 = 3
The empirical formula for the compound is FeCl3