How many grams of phosphorus react with 34.2 L of O2 at STP to form tetraphosphorus decaoxide?
according to the following formula
P4(s) + 5 O2(g) P4O10(s)
- 1 decade agoFavorite Answer
the ideal gas law is PV = nRT
at STP (T=273K, P= 1atm)
therefore at STP, there would be
PV = nRT
(1atm)(34.2L) = n(0.0821L-atm/mol.K)(273K)
n= (0.0821)(273)/34.2 moles of Oxygen gas
based from the equation, it requires 5 moles of oxygen gas to react with one mole of P4 to produce the product.
Use stoichiometric calculations,
the number of moles of P4 needed to react with (0.0821)(273)/34.2 moles of Oxygen gas is given by
[(0.0821)(273)/34.2 moles of Oxygen gas] times one mole of P4 divided by 5 moles of oxygen gas (to cancel out the moles of oxygen)
you will then arrive to the moles of phosphourous in the form P4.
convert the moles of phosphorous to its gram equivalent using the molecular wieght of P. (Take note that there are also 4 moles of P in P4).
(0.0821)(273)/34.2*1/4*(MW of P)/ 4moles of P4.
- Anonymous1 decade ago
i think the answer is 9.466? into really sure because the answer was phrased weird....i did 34.2/22.4 L to get moles....and divided 5 moles of O2 and on top, in the numerator...multiply but the ph coeffiecient...i think it is one or could be four? may be 1...so then you get 0.30535 and then mulitply molecular weight...31...and you get 9.4666g of ph....
- 1 decade ago
34.2L O2(g) * 1mol O2/22.4L O2 * 1mol P4/5mol O2 * 123.9g P4/1molP4 = 37.8g P4