need help with these QUESTION 10 (5 pts.) According to the following reaction, w

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need help with these

QUESTION 10 (5 pts.) According to the following reaction, what mass of PbCl, can form from 235 mL of 0.110 M KCI solution? Assume that there is excess Pb(NO.) 1. 2 KCI(aa) + Pb(NO').(aa) PbCl(s) + 2 KNO(aa) 7.19 g 1.30 g 3.59 g 5.94 g 1.80 g 5 points QUESTION 11 1. (5 pts.) Consider the following balanced reaction. What mass (in g) of CO, can be formed from 288 mg of O.? Assume that there is excess CH.SH present CHSH(I) + 6 O(g) 3 co,(g) + SO2(g) + 4 H,0(g) 0.209 g CO 0.126 g CO 0.198 g CO 0.792 g CO 0.396 g CO 5 points QUESTION 12 1. (5 pts.) Give the percent yield when 28.16 g of CO, are formed from the reaction of 8.000 moles of C,H with 4.000 moles of O. 2c.H. +25 Or 16 CO: + 18H20 12.50% 25.00% 20.00% 50.00%

Explanation / Answer

Q10
Moles of KCl available=0.110*0.235
We know for one mole PbCl2 we require 2 moles of KCl
Hence moles of PbCl2 =(0.110*0.235)/2
=0.012925 mol
Molar mass of PbCl2=278.1g/mol
Hence mass of PbCl2 produced=278.1*0.012925=3.59 g'

Q11

We know that for 3 moles of CO2 6 moles of O2 are required according to the equation hence if 0.288g of O2 if present then we can have
Mass of CO2=(0.288/32)*(3/6)*44=0.198g

Q12
According to the reaction we can see that O2 is limiting reagent(as for 2 moles of hydrocarbon there is requirement of 25 moles of O2 which is not fulfilled here) hence here we can see that actual production of CO2 should be (16/25)*4=2.56 mol

But the production is actually = 28.16/44=0.64 mol
hence percentage yield = (0.64/2.56)*100=25 %

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