When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water.
How many grams of sodium sulfide are formed if 1.80 of hydrogen sulfide is bubbled into a solution containing 2.40 of sodium hydroxide, assuming that the sodium sulfide is made in 94.0 yield?
How many grams of sodium sulfide are formed if 1.80 of hydrogen sulfide is bubbled into a solution containing 2.40 of sodium hydroxide, assuming that the sodium sulfide is made in 94.0 yield?
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1) write bal. rxn.: H2S + 2 NaOH ------> Na2S + 2 H2O
1 mole H2S reacts with 2 moles NaOH to form 2 moles of sodium sulfide
2) calc. moles of reactants and determine limiting reactant
H2S: 1.80 g /34 g/mole= 0.053 moles H2S
NaOH: 2.40 g/ 40 g/mole= 0.06 moles NaOH (limiting reactant)
3) 0.06 moles NaOH x 1/2 x 78 g/mole Na2S= 2.34 g Na2S theo. yield
4) 2.34 g x .94= 2.2 g Na2S
1 mole H2S reacts with 2 moles NaOH to form 2 moles of sodium sulfide
2) calc. moles of reactants and determine limiting reactant
H2S: 1.80 g /34 g/mole= 0.053 moles H2S
NaOH: 2.40 g/ 40 g/mole= 0.06 moles NaOH (limiting reactant)
3) 0.06 moles NaOH x 1/2 x 78 g/mole Na2S= 2.34 g Na2S theo. yield
4) 2.34 g x .94= 2.2 g Na2S
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