Chemistry, 22.06.2019 14:30 malenacastillo4887

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants. 4al(s)+3o2(g)→2al2o3(s) a) 1 molal, 1 mol o2 b) 4 molal, 2.6 mol o2 c) 16 molal, 13 mol o2 d) 7.4 molal, 6.5 mol o2

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Answer from: ajbyrd911

For A: Aluminium is the limiting reagent.

For B: Oxygen gas is the limiting reagent.

For C: Aluminium is the limiting reagent.

For D: Aluminium is the limiting reagent.

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

For A: 1 mol Al, 1 mol O_2

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 1 mole of aluminum will react with = $\frac{3}{4}\times 1=0.75moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

For B: 4 mol Al, 2.6 mol O_2

By stoichiometry of the reaction:

3 moles of oxygen gas reacts with 4 moles of aluminium

So, 2.6 moles of oxygen gas will react with = $\frac{4}{3}\times 2.6=3.458moles$ of aluminium.

As, the given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

Hence, oxygen gas is the limiting reagent.

For C: 16 mol Al, 13 mol O_2

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 16 mole of aluminum will react with = $\frac{3}{4}\times 16=12moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

For D: 7.4 mol Al, 6.5 mol O_2

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 7.4 mole of aluminum will react with = $\frac{3}{4}\times 7.4=5.55moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

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 :

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

Answer from: Quest

Compound with sulfer is bright red

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For the reaction shown, find the limiting reactant for each of the following initial amounts of reac...