Question

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17. How many grams of Fe2(CO3)3 (FW = 291.73) can be formed by the reaction of 65 grams FeCl3 (FW = 162.21) and 65.0 grams K2CO3 (FW = 138.21)? 2 FeCl3 + 3 K,CO® Fe2(CO3)3 + 6 KCI 18. How many grams of NO, can be produced by decomposing 721 grams N,Oz? 2 N20, Ⓡ4 NO2 + O2

Answer 1

Note

The molar mass of N2O5

Element	Symbol	Atomic Mass	# of Atoms
Nitrogen	N	14.0067	2
Oxygen	O	15.9994	5

2*14.0067+5*15.9994=108.01g/mol

The molar mass of NO2

Element	Symbol	Atomic Mass	# of Atoms
Nitrogen	N	14.0067	1
Oxygen	O	15.9994	2

1*14.0067+2*15.9994=46.0055 g/mol

Number of moles= Given mass of the substance/molar mass of the substance

Answer

The reaction is,

2FeCl₃ +3 K₂CO₃     $\rightarrow$     Fe₂ (CO₃)₃ + 6KCl.............................Balanced Chemical equation

which means,

Two moles of FeCl₃ react with three moles of K₂CO₃ will produce one mole Fe₂ (CO₃)₃ and six moles of KCl

or

For the complete reaction of 2 moles of FeCl₃, 3 moles of K₂CO₃ required

or

For the complete reaction of 2/2 moles of FeCl₃, 3/2 moles of K₂CO₃ required

_OR

For the complete reaction of 1 mole of FeCl₃, 1.5 moles of K₂CO₃ required

Here we have

65g of FeCl₃     &   65g   of K₂CO₃

Number of moles of FeCl₃ 6 5g of FeCl₃       =Given mass FeCl₃/ Molar mass of FeCl₃

                                                                             =65g /162.21g

                                                               =0.40 moles

Number of moles of K₂CO₃ 6 5g of K₂CO₃ =Given mass K₂CO₃/ Molar mass of K₂CO₃

                                                                                  =65g/138.21g

                                                              =0.47Moles

Here the reaction is between,

   0.40 moles FeCl₃ & 0.47Moles of K₂CO₃

But For the complete reaction of 1 mole of FeCl₃ , 1.5 moles of K₂CO₃ required

Hence

For the complete reaction of 1$\times$0.40 mole of FeCl₃ , 1.5$\times$0.40 moles of K₂CO₃ required

OR

For the complete reaction of 1$\times$0.40 mole of FeCl₃ , 0.60 moles of K₂CO₃ required

But we have only 0.47Moles of K₂CO_{3 ,} in this reaction amount of K₂CO₃ not sufficent for the complete reaction of 0.40 mole of FeCl_3.

At this kind of situation, K₂CO₃ known as liminiting reagent.

In this reaction only K₂CO₃ is completely consumed, so we can calculate the amount of Fe₂ (CO₃)₃ produced based on the amount of K₂CO₃ is consumed.

We know that,

Two moles of FeCl₃ react with three moles of K₂CO₃ will produce one mole Fe₂ (CO₃)₃ and six moles of KCl

OR

3 moles of K₂CO₃ will produce one 1 Fe₂ (CO₃)₃

_OR

1 moles of K₂CO₃ will produce one 1/3 moles Fe₂ (CO₃)₃

Hence,

        (1$\times$0.47) Moles of K₂CO₃ will produce one (1/3 $\times$ 0.47) moles Fe₂ (CO₃)₃

         Number of moles ofFe₂ (CO₃)₃ produced in this reaction=0.156 moles

        The amount of Fe₂ (CO₃)₃ in grams= Molar mass of Fe₂ (CO₃)₃$\times$ Number of moles

                                                         =291.73 g/mol $\times$ 0.156moes=45.70g

         The amount of Fe₂ (CO₃)₃ in grams=45.70g

Answer.2........................................................................................................................................................................................

The reaction is,

2N₂O₅    $\rightarrow$    4NO₂+ O₂

2 Moles of N₂O₅dissociate and form 4 moles NO₂

_Or

1 Moles of N₂O₅dissociate and form 2 moles NO₂

Here we have 721g of N₂O₅

Number of moles of N2O5 in 721g= Given mass of N₂O₅ / Molar mass of N₂O₅

The molar mass of N₂O₅ = 108.01 g/mol

                                                                     =721g/108.1

                                                    =6.67 moles

We know that,

1 Moles of N₂O₅dissociate and form 2 moles NO₂

Hence,

          (1 $\times$ 6.67) Moles of N₂O₅dissociate and form (2$\times$6.67) moles NO₂

The amount ofNO₂ formed in this reation=(2$\times$6.67)=13.34 moles

The amount ofNO₂ formed in this reaction in grams= Number of mole$\times$ Molar mass of NO₂

                                                                                =13.34 moles$\times$46 g/mol

The amount ofNO₂ formed in this reaction in grams      =613.61g