| (Given atomic mass : \[Fe=56,O=16,\]\[Mg=24,P=31,C=12,H=1\]) |
A) \[{{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}(g)\to 3C{{O}_{2}}(g)+4{{H}_{2}}O(l)\]
B) \[{{P}_{4}}(s)+5{{O}_{2}}(g)\to {{P}_{4}}{{O}_{10}}(s)\]
C) \[4Fe(s)+3{{O}_{2}}(g)\to 2Fe{{O}_{3}}(s)\]
D) \[2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)\]
Correct Answer: C
Solution :
\[{{C}_{3}}{{H}_{8}}(g)+5{{O}_{2}}(g)\xrightarrow[{}]{{}}3C{{O}_{2}}(g)+4{{H}_{2}}O(\ell )\] Each 1g of \[{{C}_{3}}{{H}_{8}}\]requires 3.63 g of \[{{O}_{2}}\] \[{{P}_{4}}(s)+5{{O}_{2}}(g)\xrightarrow[{}]{{}}{{P}_{4}}{{O}_{10}}(s)\] Each 1g of \[{{P}_{4}}\]requires 1.29 g of \[{{O}_{2}}\] \[4Fe(s)+3{{O}_{2}}(g)\xrightarrow[{}]{{}}2F{{e}_{2}}{{O}_{3}}(s)\] Each 1g of Fe requires 0.428 g of\[{{O}_{2}}\] \[2Mg(s)+{{O}_{2}}(g)\to 2MgO(s)\] Each 1g of Mg requires 0.66 g of \[{{O}_{2}}\] therefore least amount of \[{{O}_{2}}\]is required in option [c].
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