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bonufazy [111]
3 years ago
12

In a test of an automobile engine 1.00 L of octane (702 g) is burned, but only 1.84 kg of carbon dioxide is produced. What is th

e percentage yield of carbon?
Chemistry
1 answer:
Reptile [31]3 years ago
4 0

Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

Step 6: Calculate % yield of carbon dioxide

% yield = (actual yield / theoretical yield)*100%

% yield = (1840/2163.5)*100%

% yield = 85.05 %

The % yield of CO2 is 85.05 %

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Maple syrup has a density of 1.325 g/ml, and 100.00 g of maple syrup contains 67 mg of calcium in the form of ca2+ ions. what is
e-lub [12.9K]
Density of maple syrup = 1.325 g/ml
1000 ml contains 1325 g of maple syrup
In 100 g of maple syrup - 67 mg of Ca ions
Therefore in 1325 g of maple syrup - 67 mg /100g * 1325 g 
                                                        = 887.75 mg of Ca
this means in 1000 ml - 887.75 mg of Ca
molar mass of Ca - 40 g/mol
therefore number of moles in 1000 ml - 0.88775 g /40 g/mol
molarity of Ca - 0.022 mol/dm³
7 0
3 years ago
Read 2 more answers
Based on the information in the tables, what is the Enthalpy of the solution for each substance? Please help me with this proble
3241004551 [841]

Answer: mg(OH)2 = +37

Rbl = 31

RbOH= -83

mgl2= -144

Explanation: just did it on edge2020

5 0
3 years ago
How do you pass sience
MariettaO [177]

Answer:

Listen in Class. The first step to passing any test is to begin preparing right there in the classroom. ...

Review Lab and Lecture Notes. ...

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Explanation:

8 0
3 years ago
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kherson [118]

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

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6 0
3 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
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