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3 years ago

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O.

Chemistry

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:

14.20g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O

Step 2:

Determination of the masses of CH3(CH2)6CH3 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below

Molar Mass of CH3(CH2)6CH3 = 12 + (3x1) + 6[12 + (2x1)] + 12 + (3x1) = 12 + 3 + 6[12 + 2] + 12 + 3 = 12 + 3 + 6[14] + 12 + 3 = 114g/mol

Mass of CH3(CH2)6CH3 from the balanced equation = 2 x 114 = 228g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

From the balanced equation above, 228g of CH3(CH2)6CH3 reacted with 800g of O2 to produce 704g of CO2.

Step 3:

Determination of the limiting reactant. We must obtain the limiting reactant in order to get the maximum yield of CO2.

The limiting reactant is obtained as follow:

From the balanced equation above, 228g of CH3(CH2)6CH3 reacted with 800g of O2.

Therefore, 4.6g will react with = (4.6x800)/228 = 16.14g of O2.

The mass of O2 that reacted is lesser than the given mass ( i.e 27.4g) from the question. Therefore, CH3(CH2)6CH3 is the limiting reactant.

Step 4:

Determination of the maximum mass of the CO2 produced.

The limiting reactant is used to obtain the maximum yield in any reaction.

From the balanced equation above, 228g of CH3(CH2)6CH3 produce 704g of CO2.

Therefore, 4.6g of CH3(CH2)6CH3 will produce = (4.6 x 704)/228= 14.20g of CO2.

Therefore, the maximum mass of CO2 produced is 14.20g

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3 years ago

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3 years ago

1. 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%

Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

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