How many moles of O_2 are required to completely react with 1.77 moles of C_3H_6

Question

How many moles of O_2 are required to completely react with 1.77 moles of C_3H_6 according to the following reaction? 2C_3H_6 + 9O_2 implies 6CO_2 + 6H_2O Consider the following reaction: CH_4(g) + 4Cl_2(g) implies CCl_4(l) + 4HCL(g) How many grams of CCl_4(l) will be formed if 20.0 g of gaseous methane react with 3.50 moles of chlorine gas? The reaction of 20.0 g benzene, C_6H_6, with excess HNO_3 resulted in actually producing 15.0 g of C_6H_5NO_2. What is the percent yield? C_6H_6 + HNO_3 implies C_6H_5NO_2 + H_2O Write a balanced chemical equation (use smallest whole number coefficients) for the complete combustion of ethane, C_2H_6(g).

Explanation / Answer

12) for the reaction, 2C3H6 +9O2 -------> 6CO2 + 6H2O,

to balance this reaction ,2 moles of propene requires 9 moles of oxygen (O2) ,

therefore assume 1.77 moles of propene = 2 moles,

then how many = 9 moles of O2,

By cross multiplying ,

1.77*9=15.93/2=7.965 moles of O2,

or it can be calculated as ratio of C3H6:O2 = 2:9 means 1:4.5,

therefore 1.77*4.5=7.965 moles of O2 will be required to completeconversion of C3H6 to CO2 and H2O.

13)This reaction is balanced reaction.

CH4 (g) + 4Cl2(g) ---------->CCl4(l) + 4HCl(g)

one mole of methane will react with 4 moles of chlorine gas.

that is the ratio is 1:4 (CH4:Cl2)

20 gms=moles=20/mol.wt=20/16= 1.25 moles of methane.

If one mole of methane requires 4 moles of chlorine. then 1.25 moles of methane requires 1.25*4=5 moles of chlorine. but we have 3.5 moles of chlorine. there fore 3.5/4=0.875 moles of methane will react with chlorine gas, this means that methane is excess, and this is not the limiting reagent,the limiting reagent is chlorine.

3.5*70.90=248.15gms of chlorine (mol.wt) = 70.90

248.15*mol.wt of product / mol.wt of reactant,

= 248.15 * 153.82 / 70.90 (CCl4= mol.wt =153.82),

= 38170.43 / 70.40 ,

= 538.37 gms of CCl4.

14) Theoretical yield = gms of reactant * mol.wt of product / mol.wt of reactant ,

= 20 * 123.10 / 78.11 (mol.wt of nitro benzene = 123.10 and mol.wt of benzene = 78.11) ,

= 2462 / 78.11,

=31.51 gms ,

%practical yield = (actual yield / theoreticl yield )* 100 ,

= (15 / 31.51) * 100 ,

=0.4760 * 100,

= 47.60%

15) balanced equation for combustion of ethane (C2H6),

C2H6 + O2 --------> CO2 + H2O,

from above equation ,

C2H6 has 2 carbons and 6 hydrogens, it will give 2 CO2 and 3 H2O Molecules,

so we will require 3.5 moles of O2 (7 O atoms), now we can balanced the equation as,

C2H6 + 7/2 O2 ------> 2 CO2 + 3H2O,

OR C2H6 + 3.5 O2 --------> 2 CO2 + 3H2O,

OR 2 C2H6 + 7 O2 ------> 4 CO2 + 6 H2O.

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