Answer:
44 grams of CO₂ will be formed.
Explanation:
The balanced reaction is:
C + O₂ → CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- C: 1 mole
- O₂: 1 mole
- CO₂: 1 mole
Being the molar mass of each compound:
- C: 12 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
By stoichiometry the following mass quantities participate in the reaction:
- C: 1 mole* 12 g/mole= 12 g
- O₂: 1 mole* 32 g/mole= 32 g
- CO₂: 1 mole* 44 g/mole= 44 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
If 12 grams of C react, by stoichiometry 32 grams of O₂ react. But you have 40 grams of O₂. Since more mass of O₂ is available than is necessary to react with 12 grams of C, carbon C is the limiting reagent.
Then by stoichiometry of the reaction, you can see that 12 grams of C form 44 grams of CO₂.
<u><em>44 grams of CO₂ will be formed.</em></u>
Answer:
Explanation:
<u>1) Reactants:</u>
The reactants are:
- <em>Molecular chlorine</em>: this is a gas diatomic molecule, i.e. Cl₂ (g)
- <em>Molecular fluorine</em>: this is also a gas diatomic molecule: F₂ (g)
<u>2) Stoichiometric coefficients:</u>
- <em>One volume of Cl₂ react with three volumes of F₂</em> means that the reaction is represented with coefficients 1 for Cl₂ and 3 for F₂. So, the reactant side of the chemical equation is:
Cl₂ (g) + 3F₂ (g) →
<u>3) Product:</u>
- It is said that the reaction yields <em>two volumes of a gaseous product;</em> then, a mass balance indicates that the two volumes must contain 2 parts of Cl and 6 parts of F. So, one volume must contain 1 part of Cl and 3 parts of F. That is easy to see in the complete chemical equation:
Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)
As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.
Answer:
I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)
Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)
a) E(Pt⁺²/Fe°) = - 1.668v
b) Process is Non-spontaneous if E(cell) < 0
Explanation:
Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔
Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)
As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.
E°(Fe⁺²) = -0.44v
E°(Pt⁺²) = +1.20v
E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)
= -0.44v - (+1.20v) = - 1.64v
[Fe⁺²] = 0.0066M
[Pt⁺²] = 0.057M
n = electrons transferred = 2
E(nonstd) = E°(std) - (0.0592/n)logQ);
Q = [Pt⁺²]/[Fe⁺²]
= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v
Also, if ΔG(cell) > 0 => indicates non-spontaneous process
ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)
atomic number is equal to proton number
so the proton number will be 87
