The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
A. Sodium is correct.
Sodium is an alkali metal.
Aluminum is a post-transitional metal- still a metal but it’s character is not as metallic as sodium.
Silicon is a metalloid- it has characteristics from both the metals and the non-metals.
Phosphorous is a non-metal.
Larger gases produces more spectral lines than the smaller gases because they have more orbitals in their atoms.
Hydrogen has only one orbital in which an electron orbits. At the excited state, that is, when the electron gains energy, the number of energy level it can transcend is very few. For larger elements, they have more orbitals and when excited, they can move from the ground state to other energy levels at which they produce various unique spectral lines.
Answer: S8 + O2 → S8O2: not 100% sure
Explanation:
A balanced chemical equation happens when the quantity of the particles is required on the reactants side is equivalent to the quantity is the molecules in the items side.
