Answer:
moles of calcium hydroxide= 21.75 mol
a) 43.5 mol
b) 7.25 mol
Explanation:
Please see the attached picture for the full solution.
<span>N2 + 3H2 → 2 </span>NH3<span> from bal. rxn., 2 moles of </span>NH3<span> are formed per 3 moles of </span>H2, 2:3 moleH2<span>: 3.64 </span>g<span>/ 2 </span>g<span>/mole </span>H2<span>= 1.82 1.82 moles </span>H2<span> x 2/3 x 17
</span>
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
Chemical change or process
<h3>
Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
