In sample of gas 50.og of oxygen gas o2 take up 40l of volum.keeping the pressure constants, the amount of gas is changed the volume is 72l.how many grams of gas are now in the container?

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Answer 1 · 2018-07-09T17:41:41

$90.0 {g O}_{2}$ $90.0" g O"_2$

Assuming Pressure and Temperature are constant using the ideal gas law: $P V = n R T$ $PV=nRT$

We can say that:
$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$ $V_1/n_1=V_2/n_2$

Where V represents the volume and $n$ $n$ represents the number of mols

$50 {g O}_{2} \cdot \frac{1 {mol O}_{2}}{32 {g O}_{2}} = 1.5625 {mol O}_{2}$ $50 "g O"_2*(1 "mol O"_2)/(32 "g O"_2)= 1.5625 "mol O"_2$

$n_{2} = \frac{V_{2} \cdot n_{1}}{V_{1}}$ $n_2=( V_2*n_1)/V_1$

$n_{2} = \frac{72 L \cdot 1.5625 m o l}{40 L} = 2.8125 {mol O}_{2}$ $n_2=(72L*1.5625 mol)/(40L)=2.8125" mol O"_2$

$2.8125 {mol O}_{2} \cdot \frac{32 {g O}_{2}}{1 {mol O}_{2}} = 90.0 {g O}_{2}$ $2.8125" mol O"_2*(32 "g O"_2)/(1 "mol O"_2)=90.0" g O"_2$