Answer:Use an excess of ethane
Explanation:
The halogenation of alkanes is a substitution reaction. All the hydrogen atoms in the alkanes could be potentially substituted. How ever the reaction can be controlled by using an excess of either the alkane or the halogen. If the aim (as it is in this question) is to minimize the yield of halogenated alkanes, an excess of the alkane (in this case, ethane) is used.
According to the balanced equation of the reaction:
H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)
First, we have to get the no. of moles of H2X:
no.of moles of H2X = weight / molar mass
when we have the H2X weight = 0.1873 g & the molar mass H2X = 85 g/mol
So by substitution:
∴ no.of moles of H2X = 0.1873 /85
= 0.0022 mol
-then, we need to get no.of moles of NaOH:
from the balanced equation, we can see that 1 mol H2X = 2 mol NaOH
∴ no.of moles of NaOH = no.of moles of H2X *2
= 0.0022 * 2 = 0.0044 mol
So we can get the volume per litre from this formula:
M (NaOH) = no.of moles NaOH / Volume L
So by substitution:
0.1052 = 0.0044 / Volume L
∴Volume = 0.042 L *1000 = 42 mL
<span>Here's your balanced chemical equation:
Pb(NO3)2</span><span> (aq) + K2SO4 (aq) </span>⇒<span> PbSO4 (s) + 2KNO3 (aq)
Therefore the precipitate formed is </span>lead(II) <span>sulfate.</span>
<span>The total heat released, Q, equals number of moles, n, times molar heat of fussion, Hf. this is: Q = n * Hf. You kno Q and Hf, so you can find n as n = Q/Hf = 235.0 kJ / 12.55 kJ/mol = 18.72 mol. Now you can pass that to mass using the atomic mass of Au, which is 197 g/mol => mass = 18.73 mol * 197 g/mol = 3688.9 grams.</span>
I don't understand what you mean??
