The minimum amount of O_2,(g) consumed per gr | vedclass

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(C) To find the amount of $O_2$ consumed per gram of reactant,we calculate the mass of $O_2$ required for $1 \ g$ of each reactant:$(A)$ $C_3H_8$: Mol

Vedclass · Accepted Answer

$4Fe_{(s)} + 3O_{2(g)} 	o 2Fe_2O_{3(s)}$

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(C) To find the amount of $O_2$ consumed per gram of reactant,we calculate the mass of $O_2$ required for $1 \ g$ of each reactant:$(A)$ $C_3H_8$: Molar mass $= 3(12) + 8(1) = 44 \ g/mol$. $1 \ mol$ $C_3H_8$ reacts with $5 \ mol$ $O_2$ $(160 \ g)$. $O_2$ per gram $= 160/44 \approx 3.64 \ g$.$(B)$ $P_4$: Molar mass $= 4(31) = 124 \ g/mol$. $1 \ mol$ $P_4$ reacts with $5 \ mol$ $O_2$ $(160 \ g)$. $O_2$ per gram $= 160/124 \approx 1.29 \ g$.$(C)$ $Fe$: Molar mass $= 56 \ g/mol$. $4 \ mol$ $Fe$ $(224 \ g)$ react with $3 \ mol$ $O_2$ $(96 \ g)$. $O_2$ per gram $= 96/224 \approx 0.43 \ g$.$(D)$ $Mg$: Molar mass $= 24 \ g/mol$. $2 \ mol$ $Mg$ $(48 \ g)$ react with $1 \ mol$ $O_2$ $(32 \ g)$. $O_2$ per gram $= 32/48 \approx 0.67 \ g$.Comparing the values,the minimum amount of $O_2$ is consumed in reaction $(C)$.

The minimum amount of $O_2\,(g)$ consumed per gram of reactant is for the reaction: (Given atomic mass: $Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1$)

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