Question

What Volume of Carbon dioxide measured at 27C and 746.7 mm pressure would be obtained by treating 10.0g of pure marble with dilute hydrochloric acid?(Aq. tension at 27C is 26.7mm)​

Answer 1

${✎\huge{\blue{\texttt{\orange A\red N\green S\pink W\blue E\purple R\red࿐}}}}$

$\sf{The \:chemical \;equation \:representing\:the }$

$\sf{reaction\: is :}$

$\sf\blue{CaCO_3 + 2HCl} \longrightarrow \sf\blue{CaCl_2 + H_2O + CO_2}$

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$\huge\sf \bold \red{Step 1 :}$

$\sf{To\: calculate \:the\: volume\: of \:CO_2 \:evolved}$

$\sf{ from \: 10g \: of \: CaCO_3 \: at \: STP.}$

$\sf{ From\:the\:chemical\: equation,}$

$\sf{100g\: of \:marble\: react\; to \:produce\:22400 cm^3 }$

$\sf{ of \: CO_2 \: at \: STP}$

$\therefore \sf{10g \: of \: marble \: will \: produce \: CO_2}$

$\implies \sf\frac{22400}{100} × 10 = 2240 cm^3$

$\sf\underline\red{This \: is \: volume \: of \: CO_2 \: evolved \: at STP}$

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$\huge\sf \bold \orange{Step 2 :}$

$\sf{To \: Calculate \: the \: Volume \: of \: CO_2 \: at \: 27^o C }$

$\sf{ and \: 746.7 mm \: pressure}$

$\mathfrak{initial \: condition} \quad\quad\quad \mathfrak{final \: condition}$

$\sf{V_1 = 2240 cm^3 \quad \quad\quad\quad V_2= ? cm^3}$

$\sf{P_1 = 760 mm \quad \quad\quad\quad P_2= 746.7 - 26.7}$

$\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad \: = \sf{720 \: mm}$

$\sf{T_1 = 273K \quad \quad\quad\quad\quad T_2= 27+273}$

$\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad = \sf{300K}$

$\sf{By \: the \: gas \: equation, \: we \: have -}$

$\sf\frac{ V_2 × 720}{300} = \frac{2240×760}{273}$

$\implies \sf{ \frac{2240 \times 760}{273} \times \frac{300}{720} }$

$\implies\sf{2598.3 cm^3}$

$\underline{\boxed{\sf{Volume \: of \: Carbon \: Dioxide\frak{\pink{=2598.3 cm^3}}}}}$

Answer 2

I think the answer is 2598.3cm³ bcoz

100g of caco3 react to form 22400cm³of co2 then one gram forms 22400/100and in 10g it will be 22400/100×10=2240...so then by p1v1/t1=p2v2/t2 we will get 2598.3