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COCICONHzNH; HzOCOOCH3 COOCH; Amination of methyl terephthalic chloride to methyl 4-(aminocarbonyllbenzoate using aqueous ammonia...

Question

COCICONHzNH; HzOCOOCH3 COOCH; Amination of methyl terephthalic chloride to methyl 4-(aminocarbonyllbenzoate using aqueous ammonia

COCI CONHz NH; HzO COOCH3 COOCH; Amination of methyl terephthalic chloride to methyl 4-(aminocarbonyllbenzoate using aqueous ammonia



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Chlorine reacts with excess of ammonia to form: [P.M.T. (Kerala) 2007] (a) $\mathrm{NH}_{4} \mathrm{Cl}$ $\square$ (b) $\mathrm{N}_{2}+\mathrm{HCl}$ (c) $\mathrm{N}_{2}+\mathrm{NH}_{4} \mathrm{Cl}$ $\square$ (d) $\mathrm{N}_{2}+\mathrm{NCl}_{3}$ (e) $\mathrm{NCl}_{3}+\mathrm{HCl}$ $\square$

This question is analogous to the acid base chemistry of water and Auto protocol ISIS The auto protocol ISIS however, in this case is occurring with pure liquid ammonia rather than pure water. So instead of two waters reacting with each other to produce hydro knee um and hydroxide we're going to have to ammonia is reacting with each other, producing ammonium and um I'd where ammonium is the acid and the um it is the base. So sodium um it would serve as a base in this system. Just like if this were water, this would be our hydroxide. sodium hydroxide would be a base. If we're going to carry out a tea tray shin with sodium A might and ammonium chloride, then it will be the amid coming from sodium amid that reacts with the ammonium coming from ammonium chloride producing to ammonia. As this would be the net titillation reaction.

So the question here wants us to basically write out the balance equation for the reaction between ammonia and H three plus HCL, and that should give us and each four c l ammonium chloride, so basically ammonium chloride. So from here, we basically can see that it's all balanced out. And secondly, it wants us to calculate the number of grams of excess reaction when basically three grams of ammonia reacts with five grams of hydrochloric acid. So, first of all, we need to find the number of goals, which is mass over molar mass, which is going to be 17 and 36 point four or five, respectively. And basically from here we'll get the number of moles which is going to be is your opponent 176 and 0.137 So are limiting. Re agent here is gonna be hydrochloric acid. So, in order to determine the number of excess grams for so what, we can take the basically the difference in most so 1.76 minus this, which gives us 0.39 to 9 moles and most by that by ammonia. So that should basically be a multiple by 17 and it gives us an excess of 0.668 grams

In this problem in this problem, both both as as sin and a reason are correct and a reason is the a reason is the correct explanation. Recently, the correct explanation of given a nelson. Therefore, according to the given option option, a age correct and said part of this problem, I hope you understand the solution of this problem.

How many grams of calcium chloride can be produced, along with 62 g of ammonia, starting with 62 g of ammonia where two moles Solar masses, 17 g in one more reaction. Strike geometry. Two moles of ammonia 21 mole of calcium chloride. One mole of calcium chloride has a mass of 111 0.1 grounds. This is equal to 203 grams of calcium chloride that can be produced alongside 62 g of ammonia.


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