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Stolb23 [73]
3 years ago
9

Please help asap 18 points if help and right answer please ​

Chemistry
1 answer:
barxatty [35]3 years ago
6 0

Answer:

Lithium = 8%

Bromine = 92%

Explanation:

To calculate percent composition, you must:

- Calculate the molar mass.

- Divide the subtotal for each element's mass by the molar mass.

- Convert to a percentage

With that being said, given LiBr (Lithium Bromide), calculate the molar mass:

Lithium has an atomic weight of 7, and there is one Lithium atoms in LiBr. Bromine has an atomic weight of 80, and there is one Bromine atom in LiBr:

1(7)\\1(80)

Add the two products:

80+7

=87

The molar mass of LiBr is about 87 grams.

With that information, divide the subtotal of Lithium by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{7}{87} × 100=

8.0459%

(Round to nearest percentage):

8%

Therefore, the percent composition of Lithium in the compound Lithium Bromide is about 8%.

Now, divide the subtotal of Bromine by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{80}{87} × 100=

91.9540

(Round):

92%

Therefore, the percent composition of Bromine in the compound Lithium Bromide is about 92%.

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Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
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Answer:

NH3 is the limiting reactant

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Explanation:

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

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volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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