# How many liters of oxygen (at STP) are required to form 10.5 g of H_2O? 2H_(2(g)) + O_(2(g)) -> 2H_2O_((g))

May 21, 2015

You'd need 6.55 L of oxygen at STP to form that much water.

STP conditions, which are defined as a pressure of 100 kPa and a temperature of 273.15 K, should immediately get you thinking about molar volume of a gas.

More specifically, about the fact that 1 mole of any ideal gas occupies exactly 22.7 L at STP.

This means that all you really need to figure out is the number of moles of oxygen that took part in the reaction. If you know how many moles of oxygen are needed, you can use the molar volume of a gas at STP to calculate the volume of oxygen.

Use water's molar mass to determine how many moles of water would the reaction produce

10.5cancel("g") * "1 mole water"/(18.02cancel("g")) = "0.577 moles" ${H}_{2} O$

Now take a look at the balanced chemical equation

$2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 {H}_{2} {O}_{\left(g\right)}$

Notice the $1 : 2$ mole ratio that exists between oxygen and water. This means that 1 mole of oxygen will produce 2 moles of water.

Calculate the number of moles of oxygen by

0.577cancel("moles"H_2O) * ("1 mole "O_2)/(2cancel("moles"H_2O)) = "0.2885 moles" ${O}_{2}$

So, if 1 mole occupies 22.7 L at STP, then

0.2885cancel("moles") * "22.7 L"/(1cancel("mole")) = "6.548 L"

Rounded to three sig figs, the answer will be

${V}_{{O}_{2}} = \textcolor{g r e e n}{\text{6.55 L}}$

SIDE NOTE More often than not, you will be required to use the old definition of STP, which implies a pressure of 1 atm and a temperature of 273.15 K. Under these conditions, 1 mole of any ideal gas occupies 22.4 L.

The current definition of STP implies a molar volume of 22.7 L, but if your instructor or teacher wants you to use the old value, simply redo the final calculation using 22.4 instead of 22.7 L.