Answer :
The percentage yield of ammonium chloride of 0.81 g of N[tex]H_{4}[/tex]Cl is formed when 0.75 g of N[tex]H_{3}[/tex] is reacted with 0.6 g HCl is 89%.
How do we calculate the yield percentage of any substance?
The weight of the product that was produced compared to the theoretical yield is known as the percent yield in chemistry. In order to express the result in percent, we divide the experimental yield by the theoretical yield and multiply the result by 100.
Given that,
N[tex]H_{3}[/tex](g) + HCl(g) = N[tex]H_{4}[/tex]Cl ,
0.81 g N[tex]H_{4}[/tex]Cl,
0.75 g of N[tex]H_{3}[/tex],
0.6 g of HCl,
% of N[tex]H_{4}[/tex]Cl = ?
Molar mass of N[tex]H_{3}[/tex] = 14 + (1×3) = 17 g/mol (since N = 14, H =1)
Molar mass of HCl = 35.5 + 1 = 36.5 g/mol (since H=1, Cl=35.5)
Molar mass of N[tex]H_{4}[/tex]Cl = 14×4 + 35.5 = 53.5 g/mol
Mole of N[tex]H_{3}[/tex] = 0.75/17 = 0.0441 mol
Mole of HCl = 0.62/36.5 = 0.0169 mol
Therefore 0.0169 mol of N[tex]H_{4}[/tex]Cl will be formed since limiting reactant is HCl.
1 mol N[tex]H_{4}[/tex]Cl = 53.5 g
0.0169 mol N[tex]H_{4}[/tex]Cl = (53.5×0.0169)g
= 0.904 g
Percentage yield of N[tex]H_{4}[/tex]Cl = (N[tex]H_{4}[/tex]Cl formed / Theoretical N[tex]H_{4}[/tex]Cl product)× 100
= (0.81/0.904)×100
= 89.6%
Therefore, the yield percentage of N[tex]H_{4}[/tex]Cl is 89%.
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The complete question is :
“Ammonia Reacts With Hydrogen Chloride In The Gas Phase To Produce The Ionic Compound Ammonium Chloride Via The Reaction NH3(G) + HCl(G) ➔ NH4Cl(S). What Is The Percent Yield Of Ammonium Chloride If 0.81 G Of NH4Cl Is Formed When 0.75 G Of NH3 Is Reacted With 0.62 G HCl? 91% 34% 89% 59%”
