Answer:
We produce 2.0 moles of KCl when we have 2.00 moles K and and excess Cl2
Explanation:
Step 1: Data given
Number of moles K = 2.00 moles
Cl2 is in excess, so K is the limiting reactant
Step 2: The balanced equation
2K + Cl2 → 2KCl
Step 3: Calculate moles KCl
For 2 moles K we need 1 mol Cl2 to produce 2 moles KCl
We produce 2.0 moles of KCl when we have 2.00 moles K and and excess Cl2
Answer:
As Per Given Information
Given mass of H₂O₂ ( Hydrogen Peroxide) 4.0 gm
We've been asked to find the number of moles in Hydrogen Peroxide .
First we calculate molecular mass of Hydrogen Peroxide .
Molecular mass of H₂O₂ = 1×2 + 16 × 2
Molecular mass of H₂O₂ = 2 + 32
Molecular mass of H₂O₂ = 34 gm
Now find the number of moles in given hydrogen peroxide .
![\diamond{\boxed{\blue{\bf{Number \: of \: moles \: = \frac{Given \: mass}{Molecular \: mass} }}}}](https://tex.z-dn.net/?f=%20%20%5Cdiamond%7B%5Cboxed%7B%5Cblue%7B%5Cbf%7BNumber%20%5C%3A%20of%20%5C%3A%20moles%20%5C%3A%20%20%3D%20%20%5Cfrac%7BGiven%20%5C%3A%20mass%7D%7BMolecular%20%5C%3A%20mass%7D%20%20%7D%7D%7D%7D%20)
Put the given value we obtain
![\sf \twoheadrightarrow \: Number \: of \: moles \: = \frac{4}{34} \\ \\ \sf \twoheadrightarrow \: Number \: of \: moles \: =0.117](https://tex.z-dn.net/?f=%20%5Csf%20%5Ctwoheadrightarrow%20%5C%3A%20Number%20%5C%3A%20of%20%5C%3A%20moles%20%5C%3A%20%20%3D%20%20%5Cfrac%7B4%7D%7B34%7D%20%20%5C%5C%20%20%5C%5C%20%5Csf%20%5Ctwoheadrightarrow%20%5C%3A%20Number%20%5C%3A%20of%20%5C%3A%20moles%20%5C%3A%20%20%3D0.117)
So, the number of moles is 0.117 .
The ammonia gas, having a lower molecular weight than the hydrogen chloride, will diffuse faster and travel a greater length of the tube. Consequently, the white ring of ammonium chloride will form much closer to hydrochloric acid end of the tube. Which in conclusion your answer will be D :)
The arctic climate most likely
<u>Answer:</u> The equilibrium constant for this reaction is ![1.068\times 10^{6}](https://tex.z-dn.net/?f=1.068%5Ctimes%2010%5E%7B6%7D)
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
![Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)](https://tex.z-dn.net/?f=Ni%28s%29%2B4CO%28g%29%5Crightleftharpoons%20Ni%28CO%29_4%28g%29)
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%29%5D)
We are given:
![\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%3D-587.4kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%3D0kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%3D-137.3kJ%2Fmol)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-587.4%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%284%5Ctimes%20%28-137.3%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-38.2kJ%2Fmol)
To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:
![\Delta G^o=-RT\ln K_{eq}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K_%7Beq%7D)
where,
= Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![58^oC=[273+58]K=331K](https://tex.z-dn.net/?f=58%5EoC%3D%5B273%2B58%5DK%3D331K)
= equilibrium constant at 58°C = ?
Putting values in above equation, we get:
![-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}](https://tex.z-dn.net/?f=-38200J%2Fmol%3D-%288.314J%2FKmol%29%5Ctimes%20331K%5Ctimes%20%5Cln%20K_%7Beq%7D%5C%5C%5C%5CK_%7Beq%7D%3De%5E%7B13.881%7D%3D1.068%5Ctimes%2010%5E%7B6%7D)
Hence, the equilibrium constant for this reaction is ![1.068\times 10^{6}](https://tex.z-dn.net/?f=1.068%5Ctimes%2010%5E%7B6%7D)