Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
Remember that:
number of moles = mass/molar mass
First, we get the molar mass of the nitrogen gas molecule:
It is known the the nitrogen gas is composed of two nitrogen atoms, each with molar mass 14 gm (from the periodic table)
Therefore, molar mass of nitrogen gas = 14 x 2 = 28 gm
Second we calculate the mass of the precipitate:
we have number of moles = 0.03 moles (given)
and molar mass = 28 gm (calculated)
Using the equation mentioned before,
mass = number of moles x molar mass = 0.03 x 28 = 0.84 gm
Molar mass Argon = 39.948 g/mol
1 mol ------ 39.948 g
mol ----- 20.0 g
mol = 20.0 * 1 / 39.948
= 0.5006 moles
1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L
L = 0.5006 * 22.4
= 11.21 L
hope this helps!
Answer:
228 mL
Explanation:
M1*V1 = M2*V2
M1 = 6.58 M
V1 = ?
M2 = 3.00 M
V2 = 500 mL
V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL
