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3 years ago

Stoichiometry Please show how to get this I don't understand any part of it!!

Chemistry

1 answer:

tatiyna3 years ago

8 0

Answer:

The following is a rather lengthy discussion on chemical stoichiometry, but it will lead you through the mole concept and its application to chemical reaction stoichiometry. The problems are different than the one posted, but if you can follow this, then you can work your problem of interest.

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversions.

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation 2Na + 2H₂O => 2NaOH + H₂. ‘Reading the equation’ there are 2 moles of Na, 2 moles of water, 2 moles of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation 2Na + 2H₂O => 2NaOH + H₂, one could multiply the equation by 2 giving 4Na + 4H₂O => 4NaOH + 2H₂. The equation shows balance but is not in standard form. If one multiplies the equation by ½ gives Na + H₂ => NaOH + ½H₂. Again, the equation shows balance but is not in standard form.

Standard form of equation also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.

1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

Na + H₂O => NaOH + H₂,

Given: 1.76 x 10²⁴ atoms excess --------- ? grams

= 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

= 2.922moles produces => 2.922moles H₂ (b/c coefficients of Na & H₂ are same)

Convert moles to grams => 2.922moles H₂ x 2.000 grams H₂/mole H₂

= 5.8443 grams H₂

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again, write equation and assign values to each formula unit in the equation.

Na + H₂O => NaOH + H₂,

Given: 3.5moles excess ? grams ----

Since coefficients of balanced std equation are equal …

3.5 moles Na produces => 3.5 moles NaOH

Convert moles to grams => 3.5 moles NaOH x 40 g Na/mole Na

= 140 grams NaOH

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

Na + H₂O => NaOH + H₂

Given: excess 2.75 x 10²⁵ molecules H₂O => ? atoms NaOH ----

Convert to moles => 2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O

= 45.658 moles H₂O => 45.658 moles NaOH (coefficients are equal)

Convert moles NaOH to grams NaOH => 45.658 moles NaOH x 40 grams NaOH/mole NaOH

= 1826.33 grams NaOH

Master the mole concept and you master a lot of chemistry! Good luck.

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Answer:

Explanation:

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3 years ago

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4 years ago

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A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac

UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

$pH = pKa + log(\frac{[KF]}{[HF]})$

$pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04$

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

$\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol$

$\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles$

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

$\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles$

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)

The number of moles of HF after the reaction of KF with HCl is:

$\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles$

And the concentration of HF after the reaction of KF with HCl is is:

$C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L$

Now, from the equilibrium of equation (1) we have:

$Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}$

$Ka = \frac{x^{2}}{0.531 - x}$ (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

$pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73$

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

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4 years ago

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The correct answer is 930 grams platelets.

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So, 0.236 gallon comprise = 0.236 * 3.785 L = 0.89 L

As mentioned that 1 Liter of blood comprise 1.04 kilograms of platelets. Therefore, 0.89 L of blood will contain = 1.04 * 0.89 = 0.93 Kg platelets

1 Kg contain 1000 grams. So, number of platelets in grams will be,

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3 years ago