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3 years ago

A compound is 40.0 % C, 53.3 % 0, and 6.66 % H What is its molecular formula if its molecular mass is 60 g? Show all your work

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

7 0

The molecular mass 58%

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Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO an

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The remain gases are $o_{2}_{(g)}$ and $NO_{2}_{(g)}$

Pressure of $O_{2}_{(g)}$ $1.09 atm O_{2}_{(g)}$

Pressure of $NO_{2}_{(g)}$ $1.09 atm NO_{2}_{(g)}$

Explanation:

We have the following reaction

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}$

Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.

Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.

To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation

$PV= nRT$

We cleared the mols

$n=\frac{PV}{RT}$

PV=nrT

replace the data for each gas

Constant of ideal gases

$R= 0.082\frac{atm.l}{mol.K}$

Transform degrees celsius to kelvin

$25+273=298K$

$NO_{g}$

$n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\$

$O_{2}_{(g)$

$n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\$

Find the limit reagent by stoichiometry

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}$

$0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}$

Using $O_{2}_{(g)}$ as the limit reagent produces more $NO_{(g)}$ than I have, so oxygen is my excess reagent and will remain when the reaction is over.

$NO_{(g)}$

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}$

Using $NO_{(g)}$ as the limit reagent produces less $O_{2}_{(g)}$ than I have, so $NO_{(g)}$ is my excess reagent and will remain when the reaction is over.

Calculate the moles that are formed of $NO_{2}_{(g)}$

$2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}$

We know that for all $NO_{(g)}$ to react, 0.04 mol $O_{2}_{(g)}$ is consumed.

we subtract the initial amount of $O_{2}_{(g)}$ less than necessary to complete the reaction. And that gives us the amount of mols that do not react.

$0.086-0.04= 0.046$

The remain gases are $O_{2}_{(g)}$ and $NO_{2}_{(g)}$

calculate the volume that gases occupy

$0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }} =1.79 lNO_{2}_{(g)}$

$0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }} =1.03 l O_{2}_{(g)}$

Calculate partial pressures with the ideal gas equation

$PV= nRT$

$P=\frac{nRT}{V}$

Pressure of $O_{2}_{(g)}$

$P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}$

Pressure of $NO_{2}_{(g)}$

$P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}$

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