Hischierite (Hh) has a molar mass of 13.21 g/mol, Brattine (Bt) is 63.22 g/mol and Zachum (Zh) is 37.24 g/mol.
Here is an unbalanced reaction:
Hh2Bt+Zh2 -> HhZh+Bt3
How many grams of tribrattium (Bt3) are formed when 11.7 L of zachum gas reacts with excess dihischieritebrattide?
How many molecules of DHB (Hh2Bt) will it take to fully react with 99.64 grams of zachum gas (Zh2)?
One product of the reaction above, HhZh, eventually decomposes into Hh2and Zh2 according to the balanced equation shown below. What mass of Hh2is produced when 13.7 grams of Hh2Btreacts with excess Zh2?
2HhZh -> Hh2+Zh2
2 answers
first step is to convert all amounts to moles, so you can see the relationships.
Hh2Bt+Zh2 -> HhZh+Bt3. Let's make this less confusing.
H2B + Z2 ==> HZ + B3 and balanced that is
3H2B + 3Z2 ==> 6HZ + B3
Since Z is written as Z2 in the equation it makes me think that 1 mol Z2 will occupy 22.4 L; therefore, 11.7 L of Z2 = 11.7/22.4 = 0.522 moles.
There is 1 B3 for every 1 Z2; therefore, you will have 0.522 mols B3 formed.
Grams B3 (= grams Bt3) = mols Bt3 x 3*atomic mass Bt.
The other two are worked the same way.
I marvel at the number of permutations one can have with problems like this; I think my first take is to spend my time teaching students how to work "normal" problems with "normal elements".
H2B + Z2 ==> HZ + B3 and balanced that is
3H2B + 3Z2 ==> 6HZ + B3
Since Z is written as Z2 in the equation it makes me think that 1 mol Z2 will occupy 22.4 L; therefore, 11.7 L of Z2 = 11.7/22.4 = 0.522 moles.
There is 1 B3 for every 1 Z2; therefore, you will have 0.522 mols B3 formed.
Grams B3 (= grams Bt3) = mols Bt3 x 3*atomic mass Bt.
The other two are worked the same way.
I marvel at the number of permutations one can have with problems like this; I think my first take is to spend my time teaching students how to work "normal" problems with "normal elements".
4 answers