Chemistry, 26.11.2019 22:31 vicky10445

For each reaction below, write the balanced equation.

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Answer from: Mastersonisnot

Explanation:

Al + O₂ → AlO₂

Balanced chemical equation:

Al + O₂ → AlO₂

In this equation Aluminium is reducing agent while oxygen is oxidizing agent.

K + Br₂ → KBr

Balanced chemical equation:

2K + Br₂ → 2KBr

This is balanced chemical equation because there are equal number of potassium and bromine atoms are present on both side of equation.

Mg + HCl → MgCl₂ + H₂

Balanced chemical equation:

Mg + 2HCl → MgCl₂ + H₂

This is balanced chemical equation because there are equal number of magnesium, hydrogen and chlorine atoms are present on both side of equation.

Al + O₂ → Al₂O₃

Balanced chemical equation:

4Al + 3O₂ → 2Al₂O₃

This is balanced chemical equation because there are equal number of aluminium and oxygen atoms are present on both side of equation.

Na + MgF₂ → NaF + Mg

Balanced chemical equation:

2Na + MgF₂ → 2NaF + Mg

This is balanced chemical equation because there are equal number of sodium, magnesium and fluorine atoms are present on both side of equation.

Na + HCl → NaCl + H₂

Balanced chemical equation:

2Na + 2HCl → 2NaCl + H₂

This is balanced chemical equation because there are equal number of sodium, hydrogen and chlorine atoms are present on both side of equation.

SiO₂ + HF → SiF₄ + H₂O

Balanced chemical equation:

SiO₂ + 4HF → SiF₄ + 2H₂O

This is balanced chemical equation because there are equal number of silicon, oxygen and fluorine atoms are present on both side of equation.

FeS₂ + O₂ → Fe₂O₃ + SO₂

Balanced chemical equation:

4FeS₂ + 11O₂ → 2Fe₂O₃ +8 SO₂

Answer from: Quest

elements

explanation:

yes the other ones are right just show me a pic of ur face cuz u look hot

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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