Answer:
AgCl + H2 - Chemical Equation Balancer.
Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL

We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;

Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;

The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;

The number of grams of aluminum required to react with HCl is 0.6258 g.
Answer is Mr and S as MgS ..
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)