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.
, then the bond is non-polar.
, then the bond will be covalent.
, then the bond will be ionic.