What volume is occupied by a $64 \cdot g$ $64*g$ mass of oxygen gas confined in a piston under conditions of $STP$ $"STP"$ ?

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What volume is occupied by a $64 \cdot g$ $64*g$ mass of oxygen gas confined in a piston under conditions of $STP$ $"STP"$ ?

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Answer 1 · 2017-10-28T19:35:06

Approx. $44 \cdot L$ $44*L$ ..........

Explanation:

The molar volume of $STP$ $"STP"$ is usually quoted at $22.4 \cdot L \cdot m o l^{- 1}$ $22.4*L*mol^-1$ (these definitions vary according to the syllabus you follow....

For the molar quantity of this mass of dioxygen gas we have....

$\frac{64.0 \cdot g}{32.0 \cdot g \cdot m o l^{- 1}} = 2 \cdot m o l$ $(64.0*g)/(32.0*g*mol^-1)=2*mol$ ....

And so to find the volume occupied by the molar quantity, we take the product....

$Moles \times molar mass$ $"Moles"xx"molar mass"$ $=$ $=$ $2 \cdot m o l \times 22.4 \cdot L \cdot m o l^{- 1} = ? ? \cdot L$ $2*molxx22.4*L*mol^-1=??*L$

Note that you must simply know that ALL elemental gases SAVE the NOBLE GASES, are bimolecular, i.e. dinitrogen, dioxygen, difluorine, dichlorine....

Answer 2 · 2017-10-28T19:40:51

Use the molar volume ( $\frac{22.4 L}{mol}$ $(22.4 "L")/("mol"$ ). This value can only be used at $STP$ $"STP"$ conditions $(0^{\circ} C or 273 K, and 1 atm)$ $(0^@" C" " or " 273 " K", and 1 " atm")$ .

Explanation:

We can use dimensional analysis to convert from:

$g O_{2} \to moles of O_{2} \to L of O_{2} at STP$ $color(red)("g") " O"_2 -> color(red)(" moles ") "of" " O"_2 -> color(red)("L ")"of" " O"_2 " at STP"$

First, let's calculate the molar mass of $O_{2}$ $"O"_2$ using the Periodic Table:

$O_{2} = 2 (16.00) = 32.00 \frac{g}{mol}$ $"O"_2 = 2(16.00) = 32.00 "g"/"mol"$

Now, we are ready to begin:

$64.0 g O_{2} (\frac{1 mol O_{2}}{32.00 g O_{2}}) (\frac{22.4 L}{1 mol}) = 44.80 {L O}_{2}$ $64.0" g" " O"_2 ((1" mol" " O"_2)/(32.00" g" " O"_2))((22.4" L")/(1" mol")) = 44.80 " L O"_2$

Your final answer is therefore $44.8 {L O}_{2 (g)}$ $44.8 " L O"_(2("g"))$ with significant figures and proper units.

I hope that helps!

Answer 3 · 2017-10-28T19:45:22

$45 d m^{3}$ $45 dm^3$ rounded.

Explanation:

Molar mass of $O_{2}$ $O_2$ molecule is 31.9

Number of moles = mass in grams/ molar mass.

$\frac{64.0}{31.9} = 2.01$ $64.0/31.9= 2.01$ moles.

1 mole of of $O_{2}$ $O_2$ at S.T.P = $22.4 d m^{3}$ $22.4 dm^3$

So:

$2.01 \times 22.4 = 45.024 d m^{3}$ $2.01 xx22.4= 45.024 dm^3$

Answer 4 · 2017-10-28T19:57:51

48 $d m^{3}$ $dm^3$ OR 44.8 $d m^{3}$ $dm^3$

Explanation:

At GCSE level, I was taught that one mole of gas occupies a volume of 24 $d m^{3}$ $dm^3$ (or 24 litres). However when I searched online and learned that the accurate volume 1 mole of gas occupies is 22.4 $d m^{3}$ $dm^3$ (or 24 litres).

I will show you the calculation for both.

First we need to calculate how many moles of $O_{2}$ $O_2$ are in 64.0g using the equation: $\frac{m a s s}{m r}$ $(mass)/(mr)$ =moles

mass=64.0g
Mr( $O_{2}$ $O_2$ )= $16 \cdot 2$ $16*2$ =32

moles= $\frac{64.0}{32} = 2$ $64.0/32=2$

If 1 mole of $O_{2}$ $O_2$ = $24 d m^{3}$ $24dm^3$
2 moles of $O_{2}$ $O_2$ = $48 d m^{3}$ $48dm^3$

If 1 mole of $O_{2}$ $O_2$ = $22.4 d m^{3}$ $22.4dm^3$
2 moles of $O_{2}$ $O_2$ = $44.8 d m^{3}$ $44.8dm^3$

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What volume is occupied by a $64 \cdot g$ $64*g$ mass of oxygen gas confined in a piston under conditions of $STP$ $"STP"$ ?

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