Answer:
52.0004 grams of mass of potassium superoxide is required
Explanation:
Let moles carbon dioxide gas be n at 22.0 °C and 767 mm Hg occupying 8.90 L of volume.
Pressure of the gas,P = 767 mm Hg = 0.9971 atm
Temperature of the gas,T = 22.0 °C = 295.15 K
Using an ideal gas equation to calculate the number of moles.
n = 0.3662 mol
According to reaction, 2 moles of carbon-dioxide reacts with 4 moles of potassium superoxide.
Then 0.3662 mol of of carbon-dioxide will react with:
of potassium superoxide.
Mass of 0.7324 mol potassium superoxide:
0.7324 mol × 71 g/mol = 52.0004 g
52.0004 grams of mass of potassium superoxide is required.
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
Answer: Yes, your answer is correct.
As the warm water holds more salt then cold water the fingers cool and produce crystals of salt that soon rains down to the floor of the ocean
Geologists use a metric ruler
