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bagirrra123 [75]
2 years ago
5

How many grams sodium bromide can be formed from 51 grams of sodium hydroxide?

Chemistry
1 answer:
raketka [301]2 years ago
7 0

Explanation:

When working with moles only, you will start by applying stoichiometry to determine how the reactants will affect your amount of products in this reaction. For this question, we will assume that other reactants are in infinite qualities, so therefore, it is the amount of aluminum that we will be concerned with. You need to figure out how much aluminum is in the specified amount of aluminum bromide, and then how much aluminum hydroxide that will be able to create. Make sure all your units cancel out!

9.24 mol AlBr3 x (1 mol Al / 1 mol AlBr3) x (1 mol Al(OH)3 / 1 mol Al) = 9.24 mol AlBr3

When you're working with mole ratios that involve grams to moles conversions, the first thing you want to do is calculate the molecular weight of each component you are being asked about. Because the question was given to you as words instead of chemical formulas, you will want to figure out the chemical formulas. For example, aluminum hydroxide is Al(OH)3 and aluminum bromide is AlBr3. To calculate molecular weight, you will want to consult a periodic table, find the molecular weight for each atom, and then calculate the correct sum of each molecular weight. Make sure you keep track of the number of each atom you have, i.e. 3 oxygen and 3 hydrogen for aluminum hydroxide.

Na = 22.990 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

NaOH = 22.990 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol

Al = 26.982 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

Al(OH)3 = 26.982 g/mol + (3 x 15.999 g/mol) + (3 x 1.008 g/mol) = 78.003 g/mol

Now, if you begin with an amount of NaOH in grams, you will first have to convert that to moles in order to use the mole ratio.

24 g NaOH x (1 mol NaOH / 39.997 g NaOH) = 0.600 mol NaOH

Now, you will have to account for the part of the sodium hydroxide that will be present in the aluminum hydroxide. In this case, it is the hydroxide (OH) portion of the formula. There is one mole of OH in each mole of NaOH, but there are 3 moles of OH in each mole of Al(OH)3. You will start with the 0.600 mol NaOH you know you have and then use the mole ratio.

0.600 mol NaOH x (1 mol OH / 1 mol NaOH) x (1 mol Al(OH)3 / 3 mol OH) = 0.200 mol Al(OH)3

Finally, when you are converting from grams to grams, you will have to find the molecular weight of both the reactant and the product, convert reactants in grams to reactants in moles, then use the mole ratio, then convert the moles of product back to grams of product. This time, you are concerned about the mole ratio of sodium, as that is the element that is in both chemical formulas.

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A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

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Which statement applies to electronegativity?
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<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Electronegativity is defined as the tendency of an atom to attract the shared pair of electrons towards itself whenever a bond is formed.

This property increases as we move from left to right across a period because the number of charge on the nucleus gets increased and electrons are attracted more towards the nucleus.

This property decreases as we move from top to bottom in a group because the electrons get add up in the new shells which make them further away from the nucleus.

Thus, the correct answer is Option A.

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Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force,

Explanation:

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Answer:

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