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timurjin [86]
2 years ago
5

What is the percent yield of LiCl if I produced 30.85g LiCl and my theoretical yield was calculated to be 35.40g LiCl?

Chemistry
1 answer:
marissa [1.9K]2 years ago
7 0

Answer:

87.15%

Explanation:

To find percent yield, we can use this simple equation

\frac{Actual}{Theoretical} *100

Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.

They give us these values, so to find the percent yield, just plug the numbers in.

\frac{30.85}{35.40} *100\\\\ =87.15

So, the percent yield is 87.15%

An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.

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If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one:
Nastasia [14]

Answer:

The correct option is;

a. 255.0 g

Explanation:

The given information are;

Mass of propane, C₃H₈ in the combustion reaction = 156.06 g

The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and  four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of  (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g

3 0
3 years ago
In an oxidation half-reaction, which amount is shown?
Tatiana [17]
Hi,
The answer should be C.

Hope this helps, if you’d like further explanation please let me know.
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3 years ago
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What is responsible for storing urine
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The bladder is responsible.
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3 years ago
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The density of dry air measured at 25 Celsius is 1.19 times 10 to the negative 3 what is the volume of 50.0g of air
Oksana_A [137]

Answer:

4.2 x 10⁴ mL

Explanation:

Data Given:

Density (d) of air =  1.19 x 10⁻³g/mL

Mass of the air (m) = 50 g

Volume of the air (V) = ?

Solution:

Formula will be used

                d = m/V

As we have to find volume so rearrange the above equation

               V = m/d . . . . . . . . . . . (1)

Put values in above equation 1

                V = 50 g / 1.19 x 10⁻³g/mL

                V = 4.2 x 10⁴ mL

So,

volume of dry air = 4.2 x 10⁴ mL

4 0
2 years ago
The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 5.13 g sam
storchak [24]

Answer:

20.66 % of the ore is antimony

Explanation:

Step 1: Data given

Mass of stibnite (Sb2S3) = 5.13 grams

The Sb3+(aq) is completely oxidized by 27.7 mL of a  0.105 M aqueous solution of KBrO3(aq).

Step 2: The balanced equation

BrO3-(aq)+ 3Sb^3+(aq) + 6 H+ → Br-(aq) + 3Sb^5+(aq) + 3H2O (l)

Step 3: Calculate moles KBrO3

Moles KBrO3 = molarity * volume

Moles KBrO3 = 0.105 M *0.0277 L

Moles KBrO3 = 0.0029085 moles

Step 4: Calculate moles Bro3-

in 1 mol KBrO3 we have 1 mol K+ and 1 mol BrO3-

In 0.0029085 moles KBrO3 we have 0.0029085 moles BrO3-

Step 5: Calculate moles Sb

For 1 mol BrO3- we need 3 mol Sb^3+ to produce 1 mol Br- and 3 mol Sb^5+

For 0.029085 moles BrO3- we need 3*0.0029085 = 0.0087255 moles Sb

Step 6: Calculate mass Sb

Mass Sb = moles Sb * molar mass Sb

Mass Sb = 0.0087255 moles * 121.76 g/mol

Mass Sb = 1.06 grams

Step 7: Calculate the percentage of Sb in the ore

% Sb = (mass Sb / total mass) * 100%

% Sb = (1.06 grams / 5.13 grams) * 100 %

% Sb = 20.66 %

20.66 % of the ore is antimony

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