How many grams of phosphorus react with 36.4 L of O2 at STP to form tetraphosphorus decaoxide?
P4(s) + 5 O2(g) P4O10(s)
________g
2007-10-06 10:25:41 · 2 answers · asked by ronnie c 1 in Science & Mathematics ➔ Chemistry
1 mol = 22.4L of gas under standard conditions;
36.4L = 36.4L * (1 mol / 22.4L) = 1.625 mol;
in the equation, we have 5 molecules of O2 reacting with every 1 molecule of P4, so we need (1.625mol / 5) = 0.325mol of P4.
How many grams of P4 is 0.325mol? Since the molar mass of P is 30.974g/mol, the molar mass of P4 is 4x this (123.896g/mol).
So we need 0.325mol x 123.896g/mol = 40.2662g of P4.
Always write the units when you do these things, then you don't have to memorize equations and it's much easier to figure it out without making mistakes.
2007-10-06 10:36:20 · answer #1 · answered by zaiken 1 · 0⤊ 0⤋
Atomic weights: P=31
36.4LO2 x 1molO2/22.4LO2 x 1molP4/5molO2 x 4molP/1molP4 = 13g phosphorus
2007-10-06 17:34:52 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋