yield = 52.23 %
Explanation:
We have the following chemical reaction:
2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)
mass of aluminium = mass of bottle with aluminium pieces - bottle mass
mass of aluminium = 10.8955 - 9.8981 = 0.9974 g
mass of alum = mass of bottle with final product - bottle mass
mass of alum = 19.0414 - 9.8981 = 9.1433 g
number of moles = mass / molecular weight
number of moles of aluminium = 0.9974 / 27 = 0.03694 moles
number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles
To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:
if 2 moles of aluminium produce 2 moles of alum
then 0.03694 moles of aluminium produce X moles of alum
X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)
yield = (practical quantity / theoretical quantity) × 100
yield = (0.01929 / 0.03694) × 100
yield = 52.23 %
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reaction yield
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Answer:
Substance B, boiling point of 105 °C
Explanation:
Non volatile substances have high boiling points
1s2,2s2.2p6,3s2,3p6,3d4,4s2
Answer:
I'm so sorry if this is wrong but I think its B
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
