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How many grams of air are required to complete the combustion of 93 grams of phosphorus to produce diphosphorus pentoxide, assuming air to be 23% oxygen by mass?

Notes: All gasses under STP conditions, 1 cm^3=1 mL, 1 dm^3=1 L

2007-12-16 07:23:43 · 2 answers · asked by Anonymous in Science & Mathematics Chemistry

Using dimmensional analysis

2007-12-16 07:39:23 · update #1

2 answers

4P + 5O2>>2 P2O5
Moles P2O5 = 93 g / 141.95 g/mol = 0.655
the ratio between P2O5 and O2 is 2 : 5
Moles O2 needed = 0.655 x 5 / 2 = 1.64
Mass O2 = 1.64 mol x 32 g/mol = 52.4 g
23 : 100 = 52.4 : x
x = mass air = 227.9 g

2007-12-16 07:31:34 · answer #1 · answered by Dr.A 7 · 0 0

4P + 5O2 = 2P2O5
Molar ratios are 4:5::2
moles(P) = 93/31 = 3 (this is equivalent to 4).
moles(O2) = 3 x 5/4 = 3.75 (this is equivalent to 5)
mass(O2) = 3.75 x 32 = 120 g.

mass(air) is 120 x 100/23 = 434.8 g

2007-12-16 07:36:04 · answer #2 · answered by lenpol7 7 · 0 0

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