HERE Balance the following equations by inserting the proper coefficients.
CH4 +
102 → CO2 + H20
CaCl2 +
AgNO3 → Ca(NO3)2 +Agci
C2H60+
+
02
co2+ H20

CH4 + 2O2 ==> CO2 + 2H2O

CaCl2 + 2AgNO3 ==> Ca(NO3)2 + 2AgCl

C2H6O + 2O2 ==> 2CO2 + 3H2O

Explanation:

Important concept.

Coefficients : the big number which represents the number of molecules in an equation. we multiply the coefficient by the subscript of an element to find the number of atoms of that specific element. ex 2H3O2 To find the number of hydrogen atoms we multiply the coefficient 2 by the subscript after H 3. 3 x 2 = 6. there are 6 hydrogen atoms in 2H302

1.) CH4 + O2 ==> CO2 + H2O

First we must balance the number of hydrogen atoms on both sides. There are two hydrogen atoms on the right side but 4 hydrogen atoms on the left side. We can balance the number of hydrogen atoms by adding a coefficient of 2 to H2O.

We now have CH4 + O2 ==> CO2 + 2H2O

Next we must balance the number of Oxygen atoms. There are 2 oxygen on the right side and 4 on the left side. we can easily balance it by adding a coefficient of 2 to O2 to the right side

CH4 + 2O2 ==> CO2 + 2H2O

The equation is now balanced.

2.) CaCl2 + AgNO3 ==> Ca(NO3)2 + AgCl

==> balance # of Nitrogen and Oxygen atoms by adding a coefficient of 2 to AgNO3

CaCl2 + 2AgNO3 ==> Ca(NO3)2 + AgCl

==> balance # of Argon and Chlorine atoms by adding a coefficient of 2 to AgCl

CaCl2 + 2AgNO3 ==> Ca(NO3)2 + 2AgCl

The equation is now balanced.

3.) C2H6O + O2 ==> CO2 + H2O

==> balance the # of hydrogen atoms by adding a coefficient of 3 to H2O

C2H6O + O2 ==> CO2 + 3H2O

==> balance the number of carbon atoms by adding a coefficient of 2 to CO2

C2H6O + O2 ==> 2CO2 + 3H2O

balance the number of Oxygen atoms by adding a coefficient of 2 to O2

C2H6O + 2O2 ==> 2CO2 + 3H2O

The equation is now balanced.

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