Answer:
See explanation
Explanation:
2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)
Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles
Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles
2 moles of HCl reacts with 1 mole of calcium carbonate
x moles of HCl reacts with 0.00054 moles of calcium carbonate
x = 2 * 0.00054/1
x = 0.00108 moles of HCl
Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles
Reaction of HCl and NaOH
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Since the reaction is in the mole ratio of 1:1
0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
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The compound is sodium chloride
<span>So the oxidizing agent will receive electrons from the reducing agent and the oxidation agent will take electrons from the reducing agent.</span>
