Answer:- 27.7 grams of 
are produced.
Solution:- The balanced equation is:
let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of ![Pb(NO_3)_2 = 207.2+2(14.01)+6(16) = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%20%3D%20207.2%2B2%2814.01%29%2B6%2816%29%20%20%3D%20331.22%20gram%20per%20mol%3C%2Fp%3E%3Cp%3Emolar%20mass%20of%20NaI%20%3D%2022.99%2B126.90%20%3D%20149.89%20gram%20per%20mol%3C%2Fp%3E%3Cp%3EMolar%20mass%20of%20%5Btex%5DPbI_2)
= 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:
= 
= 
From above calculations, NaI gives least amount of , so the answer is, 27.7 g of are produced.
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Answer:
=759.95 grams.
Explanation:
The molar mass of chromium is 51.9961 g/mol
Therefore the number of moles of chromium in 156 grams is:
Number of moles =mass/RAM
=156g/51.9961g/mol
=3 moles.
From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.
Therefore 3 moles of chromium produce:
(3×2)/4 moles =1.5 moles of chromium oxide.
I mole of chromium oxide has a mass of 151.99 g
Thus 1.5 moles= 1.5mole ×151.99 g/mol
=759.95 grams.
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
