Answer:
2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)
Explanation:
An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.
In this reaction, solid lithium reacts with liquid water to form soluble lithium hydroxide and gaseous hydrogen
.
1. Molecular equation
2Li(s) + 2H₂O(ℓ) ⟶ 2LiOH(aq) + H₂(g)
2. Ionic equation
Lithium hydroxide is a soluble ionic compound, so we write it as hydrated ions.
2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)
Water or also known as H2O
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
Boyle’s Law P1V1 = P2V2
P1 = 0.80 atm V1 = 1.8 L
P2 = 1.0 atm V2 = ??
(.8 atm)(1.8 L) = (1.0 atm)(V2)
1.44 atm x L = 1 atm V2
