I need help setting up thisproblem. The 94% yield is confusing me. Can anyonehel

Question

I need help setting up thisproblem. The 94% yield is confusing me. Can anyonehelp?

When hydrogen sulfide gas is bubbledinto a solution of sodium hydroxide, the reaction forms sodiumsulfide and water. How many grams of sodium sulfide areformed if 1.50 g of hydrogen sulfide is bubbled into asolution containing 2.00 g of sodium hydroxide, assumingthat the sodium sulfide is made in 94.0 % yield?

m=? grams

Can anyone help me with this problem. I'm lost! Thanks.

Explanation / Answer

The correct balanced equation is :                H2S + 2 NaOH ----->Na2S + 2 H2O . One mole of H2S reacts with 2 moles of NaOH to give one mole ofNa2S Moles of H2S = 1.50 g / 34.08 g = 0.044 mol Moles of NaOH = 2.00 g / 40.00 g/mol = 0.05 mol . If H2S is the limiting reactant, moles of Na2S produced = 0.044 molH2S * ( 1 mole Na2S/ 1 mole H2S)                                                                                 = 0.044moles If NaOH is the limiting reactant, moles of Na2S = 0.05 mol NaOH * (1 mol Na2S / 2 mol NaOH)                                                                   = 0.025moles . Since NaOH gives less amount of product, it is the limitingreactant. 0.05 moles of NaOH will react with 0.025 moles of H2S to give 0.025moles of Na2S . Yield is 94 % So ( actual amount produced / theoretical yield) * 100 = 94 (actual amount produced / 0.025 moles ) * 100 = 94 Actual moles of Na2S produced = 0.0235 moles . Mass of Na2S produced = 0.0235 moles * 78.04 g/mol                                      = 1.83 g

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