# 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)

Relevance

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).

So,

(0.0821)(273)/34.2*1/4*(MW of P)/ 4moles of P4.

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