Science, asked by dineshrayan9031, 8 months ago

Q.8 Given the following equation: NaCl + AgNO3 → AgCl + NaNO3 How many gms of AgCl will be produced from 7.00 g of NaCl and 95.0 g of AgNO3 ?

Answers

Answered by Mihir1001

$\huge{\underline{\mathfrak{\textcolor{blue}{Answer :}}}}$

$\huge\boxed{\fcolorbox{red}{#ffdfff}{mass of AgCl obtained = 17.20 grams}}$

$\LARGE{\underline{\mathrm{\textcolor{red}{Step-by-step \: \: explanation :}}}}$

$\LARGE{\underline{\mathtt{\textcolor{violet}{Given :-}}}}$

A chemical reaction : $NaCl + AgNO_3 \longrightarrow AgCl + NaNO_3$

Mass of $NaCl = 7.00 \: g$
Mass of $AgNO_3 = 95.0 \: g$

$\LARGE{\underline{\mathtt{\textcolor{green}{To \: \: find :-}}}}$

mass of $AgCl$ produced

$\LARGE{\underline{\mathtt{\textcolor{teal}{Concept \: \: used :-}}}}$

MOLAR CONCEPT

$\LARGE{\underline{\mathtt{\textcolor{blue}{Solution :-}}}}$

From the equation, we get that reactants and products, both are in single moles.

Now,

Molar mass of $NaCl = 22.99 g + 35.45 g = 58.44 g$

$\longmapsto$ Thus, no. of moles of $NaCl$ that reacts

$\large{ \: \: \: \: \: \: \: \: \: \: = \frac{given \: mass \: of \: NaCl}{molar \: mass \: of \: NaCl} }$

$\large{ \: \: \: \: \: \: \: \: \: \: = \frac{7.00 \: g}{58.44 \: g} moles}$

$\large{ \: \: \: \: \: \: \: \: \: \: = 0.12 \: moles}$

Molar mass of $AgNO_3 = 107.87 g + 14.00 g + 3(16.00) g = 169.87 g$

$\longmapsto$ Thus no. of moles of $AgNO_3$ that reacts

$\large{ \: \: \: \: \: \: \: \: \: \: = \frac{given \: mass \: of \: AgNO_3}{molar \: mass \: of \: AgNO_3} moles}$

$\large{ \: \: \: \: \: \: \: \: \: \: = \frac{95.0 \: g}{169.87 \: g} moles}$

$\large{ \: \: \: \: \: \: \: \: \: \: = 0.56 \: moles}$

$<hr>$

We got :

No. of moles of $NaCl <$ No. of moles of $AgNO_3$

$\therefore \: , \: NaCl \: \: is \: \: the \: \: limiting \: \: reagent.$

Hence, all reactants and products will take part in the reaction with their 0.12 moles respectively.

$<hr>$

Now,

molar mass of $AgCl$

$= 107.87 g + 35.45 g$

$= 143.32 g$

Hence,

mass of $AgCl$ produced

$= 0.12 \: moles \: of \: AgCl$

$= 0.12 \: moles \times molar \: mass \: of \: AgCl$

$= 0.12 \: moles \times 143.32 g$

$= 17.1984 \: grams$

$= 17.20 \: grams$ —————[ rounding off to ( 2 decimal places ) or ( 4 significant figures ) ]

$\LARGE{\underline{\mathtt{\textcolor{magenta}{Conclusion :-}}}}$

Mass of $AgCl$ obtained $= 17.20 \: grams$

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