You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
Answer:
c and d are correct
Explanation:
In A, false because in Valence Electrons, the more the valences, the more stable an atom is.
In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.
In C, true because the more the valence electrons, the more the stability of an atom.
In D, true as electron placing is important and the reactivity of an atom is important.
So C and D are true!
I farted it tickled my butt cheeks jingled it was hot it was squishy so I did it again
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
One is through sublimation, where the mixture is heated and iodine gets converted into gaseous form, leaving behind the iron fillings.
The other is to get a magnet near the mixture and all the iron fillings get attracted to it while iodine will be left over
