The mass of AgCl obtained by the reaction of 169-8 g of AgNO3 with 29-2 g of NaCl according to following equation is 27-2 g-AgNO3-NaCl-AgCl-NaNO3 not balancedCalculate the percentage yield-Molar mass of AgNO3 is 169-8 g-mol- molar mass of NaCl is 58-4 g-mol and molar mass of AgCl is 143-3 g-mol

The balanced equation is: AgNO_3+NaCl→AgCl+NaNO_3 Moles of nbsp; Silver Chloride = 169.8169.8 = 1 mol Moles of nbsp; NaCl = 29.258.4 = 0.5 mol According t ...

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Standard XII

Chemistry

Empirical & Molecular Formula

The mass of A...

Question

The mass of $A g C l$ obtained by the reaction of $169.8 g$ of $A g N O_{3}$ with $29.2 g$ of $N a C l$ according to following equation is $27.2 g$ .
$A g N O_{3} + N a C l \to A g C l + N a N O_{3} (not balanced)$
Calculate the percentage yield.

$($ Molar mass of $A g N O_{3}$ is $169.8 g / m o l$ , molar mass of $N a C l$ is $58.4 g / m o l$ and molar mass of $A g C l$ is $143.3 g / m o l)$

33.33%

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Solution

The correct option is A 33.33%
The balanced equation is:
$A g N O_{3} + N a C l \to A g C l + N a N O_{3}$

Moles of Silver Chloride = $\frac{169.8}{169.8} = 1 m o l$
Moles of $N a C l$ = $\frac{29.2}{58.4} = 0.5 m o l$

According to the balanced chemical equation, 1 mol of Silver Nitrate requires 1 mol of $N a C l$ .
But the number of mol of $N a C l$ available is 0.5 mol.
Hence, $N a C l$ is the limiting reagent.

Therefore, the moles of Silver Chloride formed = 0.5 mol
Mass of Silver Chloride formed = $0.5 \times 143.3$ = 71.6 g

Theoretical yield = 71.6 g
Experimental yield = 27.2 g

Percentage yield
= $\frac{experimental yield}{theoretical yield} \times 100 = \frac{27.2}{71.6} \times 100 = 33.33 %$

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Similar questions

Q. The mass of

A g C l

obtained by the reaction of

169.8 g

A g N O_{3}

with

29.2 g

N a C l

according to following equation is

27.2 g

A g N O_{3} + N a C l \to A g C l + N a N O_{3} (not balanced)

Calculate the percentage yield.

(

Molar mass of

A g N O_{3}

169.8 g / m o l

, molar mass of

N a C l

58.4 g / m o l

and molar mass of

A g C l

143.3 g / m o l)

Q. What weight of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of

A g N O_{3}

?
(Molar mass of Na = 23 g/mol, Cl = 35.5 g/mol, Ag = 108 g/mol)

A g N O_{3} + N a C l \to A g C l + N a N O_{3}

. In this reaction, mass of

A g N O_{3}

and

N a C l

is equal to the mass of

A g C l

and

N a N O_{3}

. Then this chemical reaction satisfies:

Q. A sample of

N a C l O_{3}

is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be:

[Given : Molar mass of

A g C l = 143.5 g m o l^{- 1}

]

Q. The mass of

A l (O H)_{3}

that was obtained by the reaction of 133.4 g of

A l C l_{3}

with 40 g of

N a O H

according to following equation is 13 g.

A l C l_{3} + N a O H \to A l (O H)_{3} + N a C l

(not balanced)
Calculate the percentage yield.

(Molar mass of aluminium chloride is 133.34 g/mol, molar mass of

N a O H

is 40 g/mol and molar mass of aluminium hydroxde is 78 g/ mol respectively)

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