The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
The atomic mass of the element would simply be equal to
the sum of the weighted average of each isotope, that is:
atomic mass = 59.015 amu * 0.717 + 62.011 amu * (1 – 0.717)
<span>atomic mass = 59.863 amu</span>
The answer is 18.02 g/mol
Answer:
0.30 M
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t = ?
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.36 M
k is the rate constant = 0.208 s⁻¹
t = 7.30 seconds
So,
![[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D1.36%5Ctimes%20e%5E%7B-0.208%5Ctimes%207.30%7D%5C%20M%3D0.30%5C%20M)
Answer:
3920kg of lime(CaO) 1568m^3 of CO2
Explanation:
I attached the explanation!
also for calculating the volume I assumed Standard Temperature and Pressure(STP).
