Question

Identity limiting reactants (maximum product method). Consider the reaction of sodium sulfide with oxygen and water 2Na2S (aq) +20,($)+H.0(1) Na,5,0, (aq) + 2NaOH(aq) Determine the limiting reactant in a mixture containing 195 g of Nays, 72.0 g of O, and 25.0 g of H.O. Calculate the maximum maut (in gram) of sodium thiosulfate Na 5,0%, that can be produced in the reaction The limiting reactant is: 02 Nays но Amount of Na,s,o, formed

Answer 1

Balanced seation In the manner , sama 2 Nags + 204 + H2O → Nq, $203 + 2 NAOH mole of Nas - Given mass of NagS & 1959 2.5 mole molar mass of Nars 78 g/mol me more of a = 72 g 2.25 mole - 32 g/mol mole of H₂o 25 g = 1.38 mole. . 18 g/mol • According to balanced equation 2 mole of Oq react with a I mole of Na₂S and I mob of Ho. So, I mole of Or will react with ( 22 ) 1 mole of Nas and (t/ = 0.5 mole of to. Hence overall 2.25 mole of Oz will react with 2.25 mole 1 of Na₂S and (2.25 x 5 =) 1.125 mole of Ho. Now you can see Nas and H₂o are present in excess .) 1 Limiting reactant is : O₂ According to balanced equation, a mole of Og han gine. I mole of Na, S. Oz. So, I mole of Or will give ( 2 ) 0.5 mole of Na, 5 . Hence, 2.25 mole of On willy (2.25x0.5 1125 mole of Na2S Amount of Nag So formed : Male of Na, S₂0 x Malar mass of Na28203 510125 mole x 158.11 9/ mot Exact value : 177.87 gram Rounded off value = 178 gram form