The answer is A.18 let me know if you want an explanation
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 %
Argon is a suitable choice for light bulbs because it's inert. Compared to a reactive gas like oxygen, the metal filimant would burn up in a reactive enviroment, which is why a noble gas is used.
<span>The calculation of quantities in chemical equations are called Stoichiometry. Stoichiometry is a branch of chemistry which deals with relative quantities of reactants and products in chemical reactions. The correct answer is 'Stoichoimetry'. I hope this helps you. </span>
Answer:
.081 g of O2
Explanation:
4Cr + 3O2 -----> 2Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2
