Question

Aluminium oxide contains 52.90% aluminium and carbon dioxide contains 27.37% carbon. Calculate the percentage of aluminium in aluminium carbide assuming that the law of reciprocal proportion is true.​

Answer 1

$\: \: \: \: \: \: \: \: \: \:\huge{\underline{\bf{\red{Q}\pink{u}\green{e}\blue{s}\purple{t}\orange{i}\blue{o}\purple{n}}}}$

ᴀʟᴜᴍɪɴɪᴜᴍ ᴏxɪᴅᴇ ᴄᴏɴᴛᴀɪɴs 52.9% ᴀʟᴜᴍɪɴɪᴜᴍ ᴀɴᴅ ᴄᴀʀʙᴏɴ ᴅɪᴏxɪᴅᴇ 27.2% ᴄᴀʀʙᴏɴ. ᴀssᴜᴍɪɴɢ ᴛʜᴀᴛ ᴛʜᴇ ʟᴀᴡ ᴏғ ʀᴇᴄɪᴘʀᴏᴄᴀʟ ᴘʀᴏᴘᴏʀᴛɪᴏɴ ɪs ᴛʀᴜᴇ, ᴛʜᴇ ᴘᴇʀᴄᴇɴᴛᴀɢᴇ ᴏғ ᴀʟᴜᴍɪɴɪᴜᴍ ɪɴ ᴀʟᴜᴍɪɴɪᴜᴍ ᴄᴀʀʙɪᴅᴇ ᴡɪʟʟ ʙᴇ ?

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$\: \: \: \: \: \: \: \: \: \: \: \: \: {\huge{\bf{\purple{A}{\pink{n}{\blue{s}{\green{w}{\orange{e}{\red{r}}}}}}}}}$

$\large\sf \underline \red{Concept: }$

● The law of reciprocal proportions was proposed by Jeremias Ritcher in 1792.

● states that, "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."

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$\large\sf \underline \red{Calculation: }$

$\huge \mathtt \pink{Given :- }$

$\sf In \: Al₂O₃, \% \: of \: Al = 52.9\%, \% \: of \: O \\ \sf= 100 -52.9 = 47.1\%$

$\bf \green {In \: CO₂,} \\\sf\% \: of \: Carbon = 27.27\% \\ \sf \: \% of \: Oxygen = 72.73\%$

$\bf \green{In \: Al₂O₃} \\ \sf \: ∵ 52.9 \: gm \: Al \: gives \: oxygen \: after \: reaction \: \\ \sf= 72.73 gm O₂$

$\sf∴ 1 \: gm \: O₂ \: Obtained \: = \frac{52.9}{47.1} = 1.21 \approx1$

$\bf \green{So \: that \: in \: CO₂}$

$\sf∵ 27.27 \: gm \: Carbon \: gives \: 72.73 \: gm \: O₂ \: after \: reaction.$

$\sf∴ 1 \: gm \: O₂ \: Obtained \: by \\ \sf = \: \frac{27.27}{72.73} = 0.37$

$\sf{So, \: 1 \: gm \: Oxygen \: reacts \: with \: Al \: and \: C } \\ \sf \: \frac{1.12}{0.32} = 3 : 01$

$\sf∴ quantity \: of \: Al = \frac{Part \: of \: Al}{Total \: Part} \times 100 \\ \sf = \frac{3}{4} \times 100 = 75\% \approx74$

$\: \: \: \: \: \: \: \bf \purple{So, \: 74.97\% \: is \: correct.}$

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