Chemistry, 05.07.2019 20:00 CuteDoggo1828

The following data was collected when a reaction was performed experimentally in the laboratory. reaction data reactants products al(no3)3 nacl nano3 alcl3 starting amount in reaction 4 moles 9 moles ? ? determine the maximum amount of nano3 that was produced during the experiment. explain how you determined this amount.

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Answer from: linacelina6027

764.91 grams of sodium nitrate will be formed.

Explanation:

$Al(NO_3)_3+3NaCl\rightarrow AlCl_3+3NaNO_3$

Stoichiometric ratio of alluminium nirtate and sodium chloride is 1 : 3

According to the reaction one mole of aluminium nitrate reacts with 3 mole of sodium chloride.

Then 4 moles of aluminum nitrate will require : 12 mole of NaCl

According to question starting amount taken was

4 Moles of Al(NO_3)_3 and 9 mol NaCl.

If 3 moles of NaCl reacts with one mole Al(NO_3)_3

Then 9 moles of NaCl require : $\frac{1}{3}\times 9= 3 moles Al(NO_3)_3$

Hence, NaCl is considered as limiting reagent as it limits the formation of product.

So, 3 moles of Al(NO_3)_3 will react with 9 moles of NaCl and given moles of aluminium nitrate are 4, hence is present in excess and is considered as excess reagent.

9 mol NaCl to give 9 moles of NaNO_3 .

$\text{Amount of}NaNO_3=\text{number of moles of}NaNO_3\times \text{molecular weight of}NaNO_3$

Amount of $NaNO_3=9mol\times$ 84.99 g/mol = 764.91 grams

Answer from: teresasaysrawr

The maximum amount of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ formed from reaction is .

Further Explanation:

The chemical reaction that contains an equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

The steps to balance a chemical reaction are as follows:

Step1: Complete the reaction and write the unbalanced symbol equation.

In the reaction, ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ reacts with to form ${\text{NaN}}{{\text{O}}_{\text{3}}}$ and ${\text{AlC}}{{\text{l}}_{\text{3}}}$ . The unbalanced chemical equation is as follows:

${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3} + {\text{NaCl}} \to {\text{NaN}}{{\text{O}}_{\text{3}}} + {\text{AlC}}{{\text{l}}_{\text{3}}}$

Step2: Then we write the number of atoms of all the different elements that are present in a chemical reaction in the reactant side and product side separately.

On reactant side,

Number of aluminiumatoms is 1.

Number of nitrogen atoms is 3.

Number of oxygen atoms is 6.

Number of sodium atoms is 1.

Number of chlorine atoms is 1.

On product side,

Number of aluminium atoms is 1.

Number of nitrogen atoms is 3.

Number of oxygen atoms is 6.

Number of sodium atoms is 1.

Number of chlorine atoms is 1.

Step3: Initially, we try to balance the number of other atoms of elements except for carbon, oxygen, and hydrogen by multiplying with some number on any side.

To balance the number of atoms we have to multiply ${\text{NaCl}}$ by 3, ${\text{NaN}}{{\text{O}}_{\text{3}}}$ by 3. This gives,

$\boxed1\;{\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3} + \boxed3\;{\text{NaCl}} \to \boxed3\;{\text{NaN}}{{\text{O}}_{\text{3}}} + \boxed1\;{\text{AlC}}{{\text{l}}_{\text{3}}}$

Step 4: After this, we balance the number of atoms of carbon and then hydrogen atom followed by oxygen atoms, but in the reaction no carbon or hydrogen is there and oxygen is already balanced.

Step 5: So the balanced chemical equation is given as follows:

$\boxed1\;{\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3} + \boxed3\;{\text{NaCl}} \to \boxed3\;{\text{NaN}}{{\text{O}}_{\text{3}}} + \boxed1\;{\text{AlC}}{{\text{l}}_{\text{3}}}$

As per stoichiometery, the ratio of moles of ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ that react with ${\text{NaCl}}$ is 1:3. If 1 moles of ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ react with 3 moles of ${\text{NaCl}}$ , then 9 moles of ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ will react with 3 moles of ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ . Hence ${\text{NaCl}}$ is a limiting reagent.And ${\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}$ is present in excess.

So 9 moles of ${\text{NaCl}}$ will produce 9 moles of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ .

The formula to calculate the amount of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ is as follows:

${\text{amount}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}} = \left( {{\text{Number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}} \right)\left( {{\text{molecular}}\;{\text{weight}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}} \right)$

Substitute 9 for number of moles of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ , and $84.99\;{\text{g/mol}}$ for molecular weight of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ to calculate the amount of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ .

$\begin{aligned}{\text{amount}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}} &= \left( {{\text{Number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{\text{NaN}{{\text{O}}_{\text{3}}}} \right)\left( {{\text{molecular}}\;{\text{weight}}\;{\text{of}}\;{\text{NaN}}{{\text{O}}_{\text{3}}}} \right)\\&= \left( {9\;{\text{mol}}} \right)\left( {84.99\;{\text{g/mol}}} \right)\\&= 764.91\;{\text{g}}\\\end{aligned}$

The maximum amount of ${\text{NaN}}{{\text{O}}_{\text{3}}}$ formed from reaction is $764.91\;{\text{g}}$ .

Learn More:

What are the moles of oxygen formed when 58.6g of Find the enthalpy of decomposition of 1 mole of MgO:

Answer Details:

Grade: Senior School

Chapter: Chemical Reactions

Subject: Chemistry

Keywords: NaCl, Na(NO3)3, AlCl3, Al(NO3)2, 9 mol, 4 mol, stoiochiometery.

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