b. If 68.0 g of sodium hydroxide and 44.0 g of carbon dioxide are allowed to rea

Question

b. If 68.0 g of sodium hydroxide and 44.0 g of carbon dioxide are allowed to react, which is the limiting reactant and what is the maximum mass of sodium carbonate that can produced? Cao, 44.0 40.0 ND 2. 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 2.00 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 92.0% yield?

Explanation / Answer

b. Balanced Reaction is as follows:

2NaOH + CO2 --> Na2CO3 + H2O

means 2 moles NaOH and 1 mole CO2 combine to form 1 mole Na2CO3.

Initial quantities : NaOH (Molecular weight)= 40 g/mole

so, 68 g = 68g/40g/mol= 1.7 mole

CO2 (Molecular weight )= 44 g/mole

so, 44 g= 44g/44g/mol= 1 mole

We can see NaOH is the limiting reactant. 1.7 mole of NaOH will react with 0.85 mole of CO2, resulting into formation of 0.85 mole Na2CO3.

Na2CO3 Molecular Weight= 106 g/mole

so, 0.85 mole * 106g/mole= 90 grams Na2CO3 can be formed maximum.

2. Balanced Reaction is as follows:

H2S + 2NaOH --> Na2S + 2H2O

where Na2S yield is 92%, means 1 mole of H2S and 2 moles of NaOH combine to form 0.92 moles of Na2S.

H2S (MW= 34 g/mole) 2 gram = 2 g/ 34g/mol = 0.06 moles

NaOH (MW= 40 g/mol) 2 gram= 2g/ 40 g/mol = 0.05 moles

means 0.025moles H2S will react with 0.05 moles NaOH(limiting reactant) and form 0.025 * 0.92 = 0.023 moles

Mass of Na2S(MW= 78) = 0.023* 78 = 1.8 grams.

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