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stepan [7]
2 years ago
5

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 1.50 g of hydrogen sulfide is bubbled into a solution containing 1.65 g of sodium hydroxide, assuming that the limiting reagent is completely consumed ?
Chemistry
1 answer:
Mumz [18]2 years ago
5 0

Answer:

1.61 g Na₂S

Explanation:

To find the mass of sodium sulfide (Na₂S) generated from hydrogen sulfide (H₂S) and sodium hydroxide (NaOH), you need to (1) construct the balanced chemical equation, then (2) calculate the molar masses of each molecule involved, then (3) convert grams of each reagent to grams of the product (via the molar masses and mole-to-mole ratio from equation coefficients), and then (4) determine the limiting reagent and final answer. It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).

(Step 1)

The unbalanced equation:

H₂S + NaOH ---> Na₂S + H₂O

Reactants: 3 hydrogen, 1 sulfur, 1 sodium, 1 oxygen

Products: 2 hydrogen, 1 sulfur, 2 sodium, 1 oxygen

The balanced equation:

H₂S + 2 NaOH ---> Na₂S + 2 H₂O

Reactants: 4 hydrogen, 1 sulfur, 2 sodium 2 oxygen

Products: 4 hydrogen, 1 sulfur, 2 sodium, 4 oxygen

(Step 2)

Molar Mass (H₂S): 2(1.008 g/mol) + 32.065 g/mol

<u>Molar Mass (H₂S)</u>: 34.081 g/mol

Molar Mass (NaOH): 22.990 g/mol + 15.998 g/mol + 1.008 g/mol

<u>Molar Mass (NaOH)</u>: 39.998 g/mol

Molar Mass (Na₂S): 2(22.990 g/mol) + 32.065 g/mol

<u>Molar Mass (Na₂S)</u>: 78.045 g/mol

(Step 3)

1.50 g H₂S          1 mole            1 mole Na₂S          78.045 g
-----------------  x  ----------------  x  --------------------  x  -----------------  =
                           34.081 g          1 mole H₂S            1 mole

=  3.43 g Na₂S

1.65 g NaOH           1 mole              1 mole Na₂S          78.045 g
--------------------  x  ----------------  x  -----------------------  x  ----------------  =
                              39.998 g        2 moles NaOH          1 mole

=  1.61 g Na₂S

(Step 4)

Because NaOH generates less product, it will run out before all of the H₂S is used. This makes NaOH the limiting reagent and the final answer 1.61 grams Na₂S.

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