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In the reaction silver nitrate, AgNo3, with sodium chloride , NaCl, what is the theoretical yield (which it is the same of how many gram) of silver chloride, AgCl, will be produced from 85.0 g of silver nitrate when it is mixed with excess of sodium chloride? The equation for the reaction is below. Some possibly useful molar masses are as follows: AgNo3= 169.87 g/mol, AgCl=143.32 g/mol.
AgNO3(aq) + NaCl (aq) -àAgCl(s) + NaNO3(aq)
Please show calculations and write neatly

Sagot :

The theoretical yield (which it is the equal of what number of gram) of silver chloride, AgCl, may be created from eighty five.0 g of silver nitrate while it's miles blended with excess of sodium chloride is eighty two.eight %.

Calculation :

AgNO₃ +NaCl  --> AgCl + NaNO₃

m = eighty five g

M = 169.87 g mol⁻¹

n = m/M

for AgCl ,

             zero.five = m/143.2 g mol⁻¹

           m = 71.66 g

N₂O₄ + 2N₂H₄ --> 3N₂ + 4H₂O

m = forty g      3.5g

M= ninety two.02    32.05

n = zero.434     1.092

From equation it is clean that , 1 mole of N₂O₄ requrire to react with 2 moles of N₂H₄ . but we've got , less than half moles of N₂O₄ is in less ammount . so it's miles restricting react .

now ,  1.302 mole = mass/28.01 g mol⁻¹

mass of N₂ = (1.302 mol)(28.01 g mol⁻¹)

                   = 36.469 g

% yield = (real yield/ theoritical yield)*one hundred

            =(30.2g/36.469 g)*100

              = 82.8 %

Consistent with the elements of Chemical reaction Engineering guide, yield refers to the amount of a selected product formed per mole of reactant fed on.[3] In chemistry, mole is used to explain quantities of reactants and products in chemical reactions.

Learn more about yield here : https://brainly.com/question/8638404

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