Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react with excess AgNO3, all the chlorine (present as chloride) precipitates and 1.95 g of AgCl is collected. When 1.00 g of X undergoes complete combustion,0.900 g of CO2 and 0.735 g of H2O are formed. What is the empirical formula of X? [C = 12.01,
H = 1.01, N = 14.01, Cl = 35.45, Ag = 107.87].

The answer to your question is     C₁.₄ H₆ N₁Cl₁    

Explanation:

Data

Compound               CxHyNzCla

mass of the compound= 1 g

mass of AgCl = 1.95 g

mass of CO₂ = 0.900 g

mass of H₂O = 0.735 g

Empirical formula = ?

Process

1.- Calculate the moles of Chlorine

Molar weight of AgCl = 107.87 + 35.45 = 143.32 g

                                  143.32 g of AgCl 35.45 g of Cl

                                      1.95 g of AgCl  x

                                          x = (1.95 x 35.45) / 143.32

                                          x = 0.482 g of Cl

                                   35.45 g of Cl 1 mol

                                      0.482 g of Cl  x

                                       x = (0.482 x 1) / 35.45

                                       x = 0.0136 moles of Cl

2.- Calculate the moles of Carbon

Molar weight of CO₂ = 12.01 + 32 = 44.01

                                     44.01 g 12.01 g of C

                                      0.900 g x

                                       x = (0.900 x 12.01) / 44.01

                                       x = 0.246 g of C

                                      12 g of C  1 mol

                                      0.246 g of C  x

                                      x = (0.246 x 1)/12

                                      x = 0.020 moles

3.- Calculate the moles of Hydrogen

molar weight of H₂O = 18.01

                                     18.01 g of H₂O 2 g of H

                                      0.735 g of H₂O x

                                      x = (0.735 x 2) / 18.01

                                      x = 0.082 g of H

                                      1 g of H 1 mol

                                      0.082 g of H x

                                      x = (0.082 x 1) / 1

                                      x = 0.082 moles

4.- Calculate the moles of Nitrogen

Mass of Nitrogen = 1 - 0.482 - 0.246 - 0.082

                             = 0.19 g

                                      14 g of N  1 mol

                                      0.19 g of N  x

                                       x = (0.19 x 1) / 14

                                       x = 0.014

5.- Divide by the lowest number of moles

Carbon         0.020 / 0.014 = 1.4

Hydrogen     0.082 / 0.014 = 5.9

Nitrogen       0.014 / 0.014 = 1

Chlorine       0.014 / 0.014 = 1                                        

6.- Write the empirical formula

                     C₁.₄ H₆ N₁Cl₁                                                              

The empirical formula is C3H12N2Cl2

Explanation:

Step 1: Data given

When 1.00 g of X is dissolved in water and allowed to react with excess AgNO3, all the chlorine (present as chloride) precipitates and 1.95 g of AgCl is collected.

When 1.00 g of X undergoes complete combustion,0.900 g of CO2 and 0.735 g of H2O are formed.

Atomic mass of C = 12.01 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of N = 14.01 g/mol

Atomic mass of Cl = 35.45 g/mol

Atomic mass of Ag = 107.87 g/mol

Step 2: Calculate moles of Cl

Moles AgCl = Mass AgCl / molar AgCl

Moles AgCl = 1.95 grams / 143.32 g/mol

Moles AgCl = 0.0136 moles

For 1 mol AgCl we have 1 mol Cl

Moles Cl = 0.0136 moles Cl

Mass Cl = 0.0136 moles * 35.45 g/mol = 0.482 grams Cl

Step 3: Calculate moles CO2

1 g reacts with O2  to produce 0.9 g CO2  

 Moles CO2 = 0.9 grams / 44.01 g/mol = 0.02045 moles

For 1 mol CO2 we have 1 mol C

Moles C = 0.02045 moles C

Mass C = 0.02045 moles *12.01 g/mol = 0.2456 grams C

Step 4: Calculate moles H

Moles H2O = 0.735 grams / 18.02 g/mol

Moles H2O = 0.0408 moles H2O

In 1 mol H2O we have 2 moles H

Moles H = 2*0.0408 = 0.0816 moles H

Mass H = 0.0824 grams H

Step 5: Calculate mass N

Mass N = 1.00 grams - 0.482 grams Cl - 0.2456 grams C -0.0824 grams H

Mass N = 0.190 grams N

Step 6: Calculate moles N

Moles N = 0.190 grams N /14.01 g/mol

Moles N = 0.0136 moles N

Step 7: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.02045 moles / 0.0136 moles = 1.5

H: 0.0816 moles / 0.0136 moles = 6

N: 0.0136 moles / 0.0136 moles = 1

Cl: 0.0136 moles/ 0.0136 moles = 1

The empirical formula is C3H12N2Cl2

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2