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$left(mathrm{NH}_{4}ight)_{2} mathrm{CrO}_{7}$ on heating liberates a gas. The same gas will be obtained by(a) heating $mathrm{NH}_{4} mathrm{NO}_{2}$(b) heating $m...

Question

$left(mathrm{NH}_{4}ight)_{2} mathrm{CrO}_{7}$ on heating liberates a gas. The same gas will be obtained by(a) heating $mathrm{NH}_{4} mathrm{NO}_{2}$(b) heating $mathrm{NH}_{4} mathrm{NO}_{3}$(c) treating $mathrm{Mg}_{3} mathrm{~N}_{2}$ with $mathrm{H}_{2} mathrm{O}$(d) heating $mathrm{H}_{2} mathrm{O}_{2}$ on $mathrm{NaNO}_{2}$

$left(mathrm{NH}_{4} ight)_{2} mathrm{CrO}_{7}$ on heating liberates a gas. The same gas will be obtained by (a) heating $mathrm{NH}_{4} mathrm{NO}_{2}$ (b) heating $mathrm{NH}_{4} mathrm{NO}_{3}$ (c) treating $mathrm{Mg}_{3} mathrm{~N}_{2}$ with $mathrm{H}_{2} mathrm{O}$ (d) heating $mathrm{H}_{2} mathrm{O}_{2}$ on $mathrm{NaNO}_{2}$



Answers

Calculate how many moles of $\mathrm{NH}_{3}$ form when each quantity of reactant completely reacts.
$$3 \mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)$$
a. 2.6 $\mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}$
b. 3.55 $\mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}$
c. 65.3 $\mathrm{g} \mathrm{N}_{2} \mathrm{H}_{4}$
d. 4.88 $\mathrm{kg} \mathrm{N}_{2} \mathrm{H}_{4}$

So in this problem were given with this reaction equation. And we were given Ah, few Ramon's off the reactor and in which four. And we have to find out. Um, what amount off industry will be former produced from each amount off interest for So first, we're given 2.6 small interest. For so from the balanced equation, we can see that from three morph into it before we can get formal off image three. That means from goobers, see smell of injuries, for we can get to find six small times formal image three for three more English for and this will give us 3.5 more in his three. So from open six more leverage before we can get trip on five mole in history. Now we go to the next one. Inbee were given trip on five family enrage for again. They're simple as before. So from 3.55 more interest for we will get multiplied by since 3,000,000 trees forgive us, for one is three, so they could be formal in his three over three more in Bridge four, and this will give us 4.73 more in his three, then in numbers. See, we're given the mass of the reactant, which is 65.3 time in which, for so what? We have Hercules. First, you have to convert this massive number of malls, and this that's very easy. We just have to divide this mass by the molar mass of enrich for and the molar mass off interest for if we just with some of the rhetoric, mess off each for sulfur nitrogen entering masses. Ah, 14 and 14 times two is 28 plus forehead. Rose enters 1.8 times full that miss it will be, Ah 4.32 So the Mowlam, as in total, will be no. 28 class for planned 4.32 and it will be 32.3 to weaken right 32 part zero. Through to Grandpa Ramon, we're dividing the mass over molar Mass and this will give us the number of malls and it will be two point or four more interest for So you have to find out how many, most off in its three we can get from this much more off entry H for so their second. Very simple. You just have to multiply this by the back pressure. Soto and no formal enrich for times again from balanced equation for ministry supporters from Trimble interest form. So it will be formal image three over three more and bridge for And this will give us 2.72 mole in his three. So this is the number of more off any three which will get from, um do 0.0 for more a vintage for or 65.3 graham and restful. Now, in the final one, we have 4.8 Gigi off Andress four So 4.80 kg. First, we have to convert these two Graham because we'll have to divide this by the molar mass off interest for to get the number of most of interest for and seeing the molar mass is given in Graham. We need this value in Graham on Sir Ingraham. It'll be for one day. Did get demons. Um, four point dated times 1000 Graham. That means 4888 Ciro Graham. And it was for And now if we divide this by the molar mass of enriched for which have just concluded as 32 point, um 03 to Grandpa our mole. So from here we'll get says, Ah, roughly 1 52 more leverage for so for print dated Gigi and Ridge formations. 1 52 morning dress for And from here. How many? Most of any struggle get. We'll get 1.1 52 more enrich four times formal in his three over Trimmel End Ridge for, And this will give us a value off two or three more off in age three. That means from 1 52 mole of interest for our for print dated gives you of interest for will get two or three months off in his three.

Your first yes, signed oxidation numbers for each element in a direction we have federation in the three plus, you know, true and two plus is to all. Is that an oxidation number on nitrogen minus three. Hydrogen plus one Knight was in plus four oxygen minus two. Solo exhibition stare theories equal to zero in here. Also equal to zero in a Trojan in molecular form. So it is Susan state should be zero and water we have plus one minus two and Orel Oxidation state some off. Sedation state is equal to zero atom. They change here. Hate homes Day changed situation number year. Nitrogen oxidation number change Ministry 20 here minus 3 to 0. So is equal to three electrons lost and nitrogen here, plus four to Tzeitel. Bless. 4 to 0 is it will do for electrons gain. So multiply the entire equation by two. We here for in his three plus three in or two four plus three by two into plus six is too old. So the equation here we get it in his three plus six, you know, to seven into plus 12 is toe. This is balanced Redox reaction

Before me like a P is equal to K C R T Delta End K P is equal to K C Onley when Delta N is equal to zero. So for a until toe end would be equal to four moles of gas in the products minus formals of gas in the reactant, so equals zero. So for a K P is equal. The Casey for be Delta end would be equal to three moles of gas in the products subtract two moles of gas in the reactant sweats equal the one here cape. He is not equal the Casey. And for C Delta End, we have one mole of gas in the products minus one mole of gasoline. Reactant sis. And here K P is equal the Casey. So KP and Keesee are the same for equilibrium A and C.


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