Chemistry, 16.12.2019 19:31 franksjamia

4fe(s)+3o2(g)→2fe2o3(s) how many grams of fe2o3 are produced when 12.0 of iron rusts

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Answer from: krislinsanchez8088

Mass = 10.1 g

Explanation:

Given data:

Mass of Fe₂O₃ produced = ?

Mass of iron rust = 12.0 g

Solution:

Chemical equation:

4Fe + 3O₂ → 2Fe₂O₃

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 12.0 g/ 55.85 g/mol

Number of moles = 0.215 mol

Now we will compare the moles of iron and Fe₂O₃ .

Fe : Fe₂O₃

4 : 2

0.215 : 2/4×0.125 = 0.063 mol

Mass of Fe₂O₃ :

Mass = number of moles × molar mass

Mass = 0.063 mol × 159.7 g/mol

Mass = 10.1 g

Answer from: Quest

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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4fe(s)+3o2(g)→2fe2o3(s) how many grams of fe2o3 are produced when 12.0 of iron rusts...