Coal is made of 82,2% carbon, 4,6% hydrogen, 1% sulphur and the rest is incombustible. We know that 21% of air is oxygen. Calculate what volume of air will we need to burn 1 tone of this coal?

Question

Coal is made of 82,2% carbon, 4,6% hydrogen, 1% sulphur and the rest is incombustible.  We know that 21% of air is oxygen. Calculate what volume of air will we need to burn 1 tone of this coal?

Noah Aldrich · Accepted Answer

Here's my take on this one.

!! LONG ANSWER !! The idea here is that you need to treat each element's reaction with oxygen gas separately to find the number of moles of oxygen gas needed for each of the three combustion reactions. Since the conditions for pressure and temperature were not specified by the problem, I'll assume that you're at STP, Standard Pressure and Temperature. This will allow us to find the volume of oxygen gas needed by using the molar volume of a gas at STP. You can then use the fact that air is #21%# oxygen gas by volume to find the volume of air needed for the reaction. So, the three reactions that are of interest are #"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))# #color(red)(2)"H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O"_((g))# This can be rewritten as #"H"_ (2(g)) + 1/color(red)(2)"O"_ (2(g)) -> "H"_ 2"O"_ ((g))# Keep in mind that the sample contains elemental hydrogen, #"H"#, not hydrogen gas, #"H"_2#. Since you get #1/color(red)(2)# moles of oxygen gas per mole of hydrogen gas, and since one mole of hydrogen gas contains two moles of elemental hydrogen, you can say that this reaction will consume #1/4# moles of oxygen gas per mole of elemental hydrogen. Finally, you have #"S"_ ((s)) + "O"_ (2(g)) -> "SO"_ (2(g))# These reactions tell you that you need Use the given percent composition to find how much carbon, hydrogen, and sulfur you get in that sample of coal #"For C: " 1 color(red)(cancel(color(black)("tonne coal"))) * "82.2 tonnes C"/(100color(red)(cancel(color(black)("tonnes coal")))) = "0.822 tones C"# #"For H: " 1 color(red)(cancel(color(black)("tonne coal"))) * "4.6 tonnes H"/(100color(red)(cancel(color(black)("tonnes coal")))) = "0.046 tones H"# #"For S: " 1 color(red)(cancel(color(black)("tonne coal"))) * "1 tonne S"/(100color(red)(cancel(color(black)("tonnes coal")))) = "0.010 tones S"# Convert these to grams by using the conversion factor #color(purple)(|bar(ul(color(white)(a/a)color(black)("1 tonne" = 10^3"kg" = 10^6"g")color(white)(a/a)|)))# You will have #"For C: " 0.822 color(red)(cancel(color(black)("tonnes"))) * (10^6"g")/(1color(red)(cancel(color(black)("tonne")))) = 8.22 * 10^5"g C"# #"For H: " 0.046 color(red)(cancel(color(black)("tonnes"))) * (10^6"g")/(1color(red)(cancel(color(black)("tonne")))) = 4.6 * 10^4"g H"# #"For S: " 0.010 color(red)(cancel(color(black)("tonnes"))) * (10^6"g")/(1color(red)(cancel(color(black)("tonne")))) = 1.0 * 10^4"g S"# Next, use the molar masses of the three elements to determine how many moles of each you have in your sample #"For C: " 8.22 * 10^5color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = 6.84 * 10^4"moles C"# #"For H: " 4.6 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 4.56 * 10^4"moles H"# #"For S: " 1.0 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = 3.12 * 10^2"moles S"# Use the aforementioned mole ratios to find the number of moles of oxygen gas needed for each individual reaction #6.84 * 10^4color(red)(cancel(color(black)("moles C"))) * "1 mole O"_2/(1color(red)(cancel(color(black)("mole C")))) = 6.84 * 10^4"moles O"_2# #4.56 * 10^4color(red)(cancel(color(black)("moles H"))) * (1/4color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole H")))) = 1.14 * 10^4"moles O"_2# #3.12 * 10^2color(red)(cancel(color(black)("moles S"))) * "1 mole O"_2/(1color(red)(cancel(color(black)("mole S")))) = 3.12 * 10^2"moles O"_2# The total number of moles of oxygen gas needed will be equal to #n_(O_2) = 6.84 * 10^4"moles" + 1.14 * 10^4"moles" + 3.12 * 10^2"moles"# #n_(O_2) = 8.0012 * 10^4"moles O"_2# Now, STP conditions are currently defined a pressure of #"100 kPa"# and a temperature of #0^@"C"#. Under these conditions for pressure and temperature, one mole of any ideal gas occupies a volume of #"22.7 L"#. #color(blue)(|bar(ul(color(white)(a/a)"1 mole" = "22.7 L"color(white)(a/a)|))) -># the molar volume of a gas at STP Use this to find the volume of oxygen gas needed for the reaction #8.0112 * 10^4 color(red)(cancel(color(black)("moles O"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole O"_2)))) = 1.82 * 10^6"L"# Finally, air is said to be #21%# oxygen gas by volume, i.e. every #"100 L"# of air will contain #"21 L"# of oxygen gas. This means that you need #1.82 * 10^6 color(red)(cancel(color(black)("L O"_2))) * "100 L air"/(21color(red)(cancel(color(black)("L O"_2)))) = color(green)(|bar(ul(color(white)(a/a)8.7 * 10^6"L air"color(white)(a/a)|)))# I'll leave the answer rounded to two sig figs. SIDE NOTE More often than not, the molar volume of a gas at STP is given as #"22.4 L"#. This value corresponds to the old definition of STP conditions, which imply a pressure of #"1 atm"# and a temperature of #0^@"C"#. If this is the value that was given to you, simply redo the calculations using #"22.4 L"# instead of #"22.7 L"#.

Mateo Aguilar · Answer

undefined To calculate the volume of air needed to burn 1 ton of coal, you can use the stoichiometry of the combustion reaction. First, determine the moles of carbon and hydrogen in the coal based on the given percentages. Then, use the balanced combustion equation to find the moles of oxygen required. Finally, convert the moles of oxygen to volume using the ideal gas law.