It is effected by diffusion (the power of smell and wind spread) but a solid is not.
Answer:
207g
Explanation:
Given compound:
Ca(ClO₃)₂
The equivalent mass can be derived by summing the molar masses of each atom
Molar mass of Ca = 40
Cl = 35.5
O = 16
Now solve;
Molar mass = 40 + 2(35.5 + 3(16)) = 207g
The equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.
<h3>What is the equilibrium expression?</h3>
The equilibrium expression shows how much of the reactants are converted into products. If the equilibrium constant is large and positive, most of the reactants have been converted into products.
Thus, the equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.
Learn more about equilibrium constant:brainly.com/question/10038290
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Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?
mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>