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tensa zangetsu [6.8K]
3 years ago
15

Many home barbeques are fueled with propane gas (C3H8)(C3H8). Part A What mass of carbon dioxide is produced upon the complete c

ombustion of 27.9 LL of propane (the approximate contents of one 5-gallon tank)
Chemistry
1 answer:
Andrew [12]3 years ago
3 0

Answer:

41264 g of CO₂

Explanation:

Combustion reaction is:

C₃H₈ + 5O₂ →  3CO₂  +  4H₂O

1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.

In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.

We have the volume of propane but we need moles of it, so we need to apply density.

Density = mass / volume so mass = density . volume.

Density of propane is: 493 g/L

Mass of propane is 493 g/L . 27.9L = 13754.7 g

We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles

According to reaction, 1 mol of propane can produce 3 moles of CO₂

Our 312.6 moles will produce 312.6 . 3 = 937.8 moles

We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g

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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.

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