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IrinaVladis [17]
3 years ago
13

Sodium metal (Na) reacts explosively with water to form sodium hydroxide (NaOH) and hydrogen gas. How many moles of sodium are n

eeded to produce 154 moles of hydrogen gas? (Hint: Balance the equation.)
Chemistry
1 answer:
Sati [7]3 years ago
7 0

Answer:

308 moles of sodium

Explanation:

The balanced equation for the chemical reaction between sodium metal (Na) and water (H₂O) is the following:

2 Na(s) + 2 H₂O → 2 NaOH(aq) + H₂(g)

From the equation, we can see that 2 moles of Na react with 2 moles of H₂O to give 2 moles of NaOH and 1 mol of H₂ (hydrogen gas). So the stoichiometric mole ratio between Na and H₂ is: 2 mol Na/1 mol H₂. Thus, we multiply the mole ratio by the moles of H₂ to be produced to obtain the moles of Na required:

moles of Na required =  2 mol Na/1 mol H₂ x 154 moles H₂ = 308 moles Na

Therefore, 308 moles of sodium are needed to produce 154 moles of hydrogen gas.

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How many moles of zinc, Zn, are in 1.31×1024 Zn atoms?
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Explanation:

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2 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
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Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

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Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

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M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
Consider the reaction below. HI + H2O mc014-1.jpg H3O+ + I– Which is an acid-conjugate base pair? HI and H2O H2O and H3O+ H3O+ a
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Answer: Option (d) is the correct answer.

Explanation:

According to Bronsted-Lowry, species which donate a proton are known as acid. The species which accept a proton are known as a base.

In the given reaction, acids and bases are as follows.

    HI  +   H_{2}O \rightarrow    H_{3}O^{+}           +         I^{-}

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Therefore, the acid HI loses a proton to form a conjugate base that is I^{-}.

Thus, we can conclude that HI and I^{-} is an acid conjugate base pair.

4 0
3 years ago
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