# Question edeee

Jan 20, 2015

You'd need $\text{24.6 mL}$ of oxygen for the combustion of $\text{CO}$ and $\text{10.0 mL}$ of oxygen for the combustion of hydrogen.

Let's start with the combustion of carbon monoxide. The balanced chemical equation is

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

Since no mention of pressure or temperature is made, I'll assume we're at STP, where pressure is equal to 1 atm and temperature is equal to 273.15 K. Moreover, at STP, 1 mole of any ideal gas occupies 22.4 L, or ${\text{22.4 dm}}^{3}$. Since you have ${\text{50 cm}}^{3}$ of carbon monoxide, the number of moles you have is

$n = \frac{V}{V} _ \left(\text{molar") = (50 * 10^(-3)L)/("22.4 L/mol}\right) = 0.0022$ $\text{moles}$

Notice that you have a $2 : 1$ mole ratio between $\text{CO}$ and ${\text{O}}_{2}$, which means that the number of oxygen moles you have is

"0.0022 moles CO" * ("1 mole O"_2)/("2 moles CO") = 0.0011 $\text{moles}$

This translates into a volume of

V = n_("oxygen") * V_("molar") = "0.0011 moles" * 22.4 "L"/"mole"

$V = \text{0.0246 L} = 24.6$ $\text{mL}$

This is the balanced chemical equation for the combustion of hydrogen

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

The same strategy applies for this reaction as well. The number of hydrogen moles is

$n = \frac{V}{V} _ \left(\text{molar") = (20*10^(-3)"L")/("22.4 L/mol}\right) = 0.00089$ $\text{moles}$

Once again, we have a $1 : 1$ mole ratio to apply, so the number of oxygen moles is

"0.00089 moles H"_2 * ("1 mole O"_2)/("2 moles H"_2) = 0.00045 $\text{moles}$

Therefore, the volume of oxygen needed is

V = n_("oxygen") * V_("molar") = "0.00045 moles" * 22.4 "L"/"mole"#

$V = \text{0.010 L} = 10.0$ $\text{mL}$