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irakobra [83]
3 years ago
8

What mass of carbon dioxide is produced upon the complete combustion of 26.5 L of propane (the approximate contents of one 5-gal

lon tank)? Assume that the density of the liquid propane in the tank is 0.621 g/mL. (Hint: Begin by writing a balanced equation for the combustion reaction.)
Chemistry
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

mCO2= 49.6932 kgCO2

Explanation:

Hello! Let's solve this!

First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O

We see that each mole of C3H8 (propane) we get 3 moles of CO2

From the propane volume we can obtain the grams of propane used.

molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane

mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2

mCO2= 49.6932 kgCO2

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                                     Ag₂S  +  HCl     →     AgCl  +  H₂S

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                247.8 g (1 mol) of Ag₂S produces  =  143.32 g (1 mol) of AgCl

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Solving for X,

                      X  =  (84.3 g × 143.32 g) ÷ 247.8 g

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<u>Calculate amount of H</u><u>₂</u><u>S produced</u><u>,</u>

According to equation,

                247.8 g (1 mol) of Ag₂S produces  =  34.1 g (1 mol) of H₂S

So,

                         84.3 g of Ag₂S will produce  =  X g of H₂S

Solving for X,

                      X  =  (84.3 g × 34.1 g) ÷ 247.8 g

                      X  =  11.60 g of H₂S

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