posted by Lisa .
Determine if HNO3 can dissolve in the following metal:
1) Write a balanced chemical reaction showing how the 2.09g Cu dissolves in HNO3.
I had put down: Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
but it's wrong.
2) Determine the minimum volume of 6.3M HNO3 required to completely dissolve the sample (in mL).
If you posted the problem as written then it is not written very well. Your equation looks ok to me for concentrated HNO3. For dilute HNO3 it is
3Cu + 8HNO3==> NO etc. Perhaps the prof want dilute HNO3 used.
mols Cu = grams/molsr mass
Convert mols Cu to mols HNO3 using the coefficients in the balanced equation.
Then convert mols HNO3 to grams. g = mols x molar mass.
I worded the question exactly like they have asked it.
What would be the equation for HNO3 with Sn? And how do you figure out these equations?
In response to the HNO3, Cu reacts with concd HNO3 to produce NO2. Cu reacts with dilute HNO3 to produce NO. Therefore, unless the problem differentiates for you, you are left to guess which the question wants.
Sn is above H in the activity series. With HCl it forms H2 gas and SnCl2. I would expect with HNO3 and you would get more SnO2 and NO2 since HNO3 is such a strong oxidizing agent.Pb would react similarly The figuring out part comes in several waysI think.
1. The general rules (metals above H in the activity series displace H2 gas and form the salt. Metals below H don't react with most acids.
2. HNO3 is both an acid and an oxidization agent. Metals above H react with HNO3 to form, in most cases, the higher oxidation state of the metal nitrate or oxide and NO or NO2 depending upon concn HNO3 used. I suspect some H2 gas is mixed in. With STRONG reducing agents (Al and those other metals at the top of the series HNO3 is reduced all the way down to NH3.
3. With metals below H, NO and NO2 are the primary products + the nitrate or oxide. All of that depends upon the concn of HNO3 used.
4. Finally, Pb, Sn, and those metals closely allied with these two in the periodic table produce many complex ions along with reactions that are out of the ordinary.
5. Then experience helps.
I know this isn't very definitive but the truth is that it takes many years in the business of this specific part in chemistry to absolutely always know what will happen.