One of the biggest dangers of storms that produce heavy rains or storm surges is __

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many

Mumz [18]

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

Molar Mass (H₂S): 34.081 g/mol

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

Molar Mass (NaOH): 39.998 g/mol

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

Molar Mass (Na₂S): 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.

5 0

2 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

Write a balanced chemical equation for the incomplete combustion of gaseous ethane (C2H6). What is the sum of the coefficients i

Delicious77 [7]

Answer:

C2H6 +02 ===>Co2 + H20.

Explanation:

7 0

3 years ago

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻ (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

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