5.0 g of iron is reacted with 5.0 g of water according to the chemical equation

Question

5.0 g of iron is reacted with 5.0 g of water according to the chemical equation shown below. Which one of the following statements is false? Fe(s)+ H_2O(l) rightarrow Fe_3O_4(s) + H_2(g) 6.91 g of Fe_3O_4 are produced. 2.85 g of H_2O are left over. Mass is conserved in this reaction. Water is the limiting reactant. When 7.00 times 10^22 molecules of ammonia react with 6.00 times 10^22 molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced? NH_3(g) + O_2(g) rightarrow N_2(g) + H_2O(g) 1.63 g 1.86 g 4.19 g 6.51 g If the percent yield for the following reaction is 75.0%, and 45.0 g of NO_2 are consumed in the reaction, how many grams of nitric acid, HNO_3(aq) are produced? 3 NO_2(g) + H_2O(l) rightarrow 2 HNO_3(aq) + NO(g) 30.8 g 41.1 g 54.8 g 69.3 g If the density of ethanol, C_2H_5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO_2 according to the following chemical equation? C_2H_5OH(l) + O_2(g) rightarrow CO_2(g) + H_2O(l) 6.19 mL 9.95 mL 19.9 mL 39.8 mL

Explanation / Answer

Atomic weight of Fe= 56, molar mass of water= 18, molar mass of Fe3O4= 56*3+64= 232

The reaction is 3Fe+4H2O-->Fe3O4 + 4H2

3 moles of Fe requires 4 moles of water. So molar ratio of Fe: H2O= 3:4= 1:1.33

Moles of Fe= mass/atomic weight= 5/56= 0.0892

Moles of water= 5/18= 0.28

Actual molar ratio ( Actual ) of Fe: H2O =0.0892: 0.28 = 1:3.13

Since water is more than the required ratio, it is excess. Hence, water is limiting

3 mole of Fe gives 1 mole of Fe3O4

0.0892 moles of Fe gives 0.0892*1/3= 0.0297

Mass of Fe3O4= 232*0.02976=6.89 gm

Moles of H2O consumed = (4/3)*0.0892= 0.1189

Mole s of water consumed =18*0.1189 = 2.14 gm

Mass of water remains =5-2.14= 2.86 gm

Law of conservation is always obeyed .Hence only false statement is D.

2.

6.023*1023 molecules are there in 1 mole of any substance

Accordingly

Moles : NH3= 7*1022/6.023*1023 =0.116, Oxygen = 6*1022/(6.023*1023)=0.0996

The balanced reaction is 2NH3+1.5O2-->N2 + 3H2O

2 moles of NH3 gives 1 mole of N2.

Molar ratio of reactants NH3: O2 ( theoretical) =2: 1.5 = 2/1.5 : 1 =1.33 : 1

Actual molar ratio of reactants :   0.116: 0.0996 = 0.116/0.0996 :1=1.16 :1

So limiting reactants is NH3.

Moles of N2 is limited by NH3. Hence moles of N2= 0.116/2= 0.058

Mass of N2= moles* molar mass = 0.058*28= 1.624 gm ( A is correct)

3.

Molar mass : NO2= 14+32= 46 , H2O= 18, HNO3= 63, NO= 30

Mole of NO2= mass/molar mass = 45/46= 0.9782

The balanced reaction is 3NO2+H2O--> 2HNO3 + NO

3 moles of NO2 when completly converted gives 2 moles of HNO3

0.9782 moles of NO2 gives 0.9782*2/3= 0.6521 HNO3

But yields is only 75%. , moles of HNO3 formed = 0.6521*0.75=0.489 moles

Mass of HNO3= moles* molar mass = 0.489*63= 30.8 gm ( A is correct)

4.

The balances reaction is C2H5OH + 3O2--> 2CO2 + 3H2O

Mollar masses : C2H5OH= 46, CO2= 44

Moles = mass/molar mass

Moles of CO2= 15/44 = 0.341

Moles of   C2H5 OH required =0.5 times CO2= 0.5* 0.341= 0.1705

Mass of C2H5OH= 0.1705*46=7.843 gm

Volume= mass/density = 7.843/0.789=9.94 ml ( B is correct)

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