Consider the following graph, which is plotted between the mass of N a 2 [ Fe ( CN ) 5 NO ] taken (with a fixed mass of N a 2 S taken) and the mass of the product N a 4 [ Fe ( CN ) 5 NOS ] f

Question

Consider the following graph, which is plotted between the mass of $N a_{2} [Fe (CN)_{5} NO]$ taken (with a fixed mass of $N a_{2} S$ taken) and the mass of the product $N a_{4} [Fe (CN)_{5} NOS]$ formed. If a total of 702 g is present at the end of the reaction with the same amount of $N a_{2} S$ taken as used in the above observation, find the number of moles of $N a_{2} [Fe (CN)_{5} NO]$ taken in the beginning.

(Given: Molar masses of $N a_{2} S$ = 78 g; $N a_{2} [Fe (CN)_{5} NO]$ = 312 g; $N a_{4} [Fe (CN)_{5} NOS]$ = 390 g)

TestHub · Accepted Answer

The given graph shows moles of Na 2 ​ [ Fe ( CN ) 5 ​ NO ] that can react to a maximum amount = 1 mole So, the moles of Na 2 ​ S reacting with 1 mole of Na 2 ​ [ Fe ( CN ) 5 ​ NO ] is also 1 mole. Let with 1 mole of Na 2 ​ S 1 ​ ′ n ′ mole of Na 2 ​ [ Fe ( CN ) 5 ​ NO ] is taken in the beginning so that a total of 702 g is present in the end. N a 2 ​ S + N a 2 ​ [ Fe ( CN ) 5 ​ NO ] → N a 4 ​ [ Fe ( CN ) 5 ​ NOS ] 1 mole n mole - -n-1 1 702 = (n-1) 312 +390 =312n +78 n = 2

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