What coefficients are needed to balance the equation for the complete combustion of methane? Enter the coefficients in the order CH4CH4, O2O2, CO2CO2, and H2OH2O, respectively. Express your answer as four integers, separated by

The balanced equation is given below:

CH4 + 2O2 —> CO2 + 2H2O

The coefficients are: 1, 2, 1, 2

Explanation:

The equation for the Combustion of methane (CH4) is given below:

CH4 + O2 —> CO2 + H2O

Now we can balance the equation above as follow:

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

CH4 + O2 —> CO2 + 2H2O

Now, there are a total of 4 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 2 in front of O2 as shown below:

CH4 + 2O2 —> CO2 + 2H2O

Now the equation is balanced.

The coefficients are: 1, 2, 1, 2

The coefficients for CH4; O2; CO2; H2O  are 1; 2; 1; 2

Explanation:

Step 1: Data given

The reactants are : CH4 and 02

The products are CO2 and H2O

Step 2: The unbalanced equation

CH4 + O2 → CO2 + H2O

Step 3: Balancing the equation

CH4 + O2 → CO2 + H2O

On the left side we have 4x H (in CH4) on the right side we have 2x H in (H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2.

CH4 + O2 → CO2 + 2H2O

On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in H2O). To multiply the amount of O on both side we have to multiply O2 on the left side by 2. Now the equation is balanced.

CH4 + 2O2 → CO2 + 2H2O

The coefficients for CH4; O2; CO2; H2O  are 1; 2; 1; 2

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g