Question

Balance
the
following chemical equations.
NaOH + H2SO4+ Na2SO4+ H2O.​

Answer 1

Answer:

$\mathbb{GIVEN \: REACTION :-}$

$NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + H_{2} O_{(l)}$

-   -    -    -    -   -    -    -     -    -    -    -    -    -    -   -   -   -    -   -   -   -   -   -   -   -   -

$\mathbb{TO\: BALANCE :-}$

Steps followed while balancing an equation :-

1. Write the unbalanced equation to show the reactants and products.
2. Write down how many atoms of each element there are on each side of the reaction arrow.
3. Add coefficients (the numbers in front of the formulas) so the number of atoms of each element is the same on both sides of the equation. It's easiest to balance the hydrogen and oxygen atoms last.
4. Indicate the state of matter of the reactants and products and check your work.

Step 1 :-

$NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + H_{2} O_{(l)}$

Step 2 :-

$\boxed{\begin{minipage}{15 em}\textbf{LHS OF THE REACTION} \\ \\Na atoms = 1 \\O atoms = 1 + 4 = 5 \\H atoms = 1 + 2 = 3 \\S atoms = 1 \\\end{minipage}} \quad \boxed{\begin{minipage}{15 em}\textbf{RHS OF THE REACTION}\\ \\Na atoms = 2 \\O atoms = 4 + 1 = 5 \\H atoms = 2 \\S atoms = 1 \\\end{minipage}}$

[LHS = Left Hand Side &

RHS = Right Hand Side]

Step 3 :-

Na atoms

• on the LHS = 1
• on the RHS = 2

Thus, adding coefficient "2" in front of NaOH on the LHS

$2NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + H_{2} O_{(l)}$

Now in LHS,

• Na atoms = 2
• O atoms  = 2 + 4 = 6
• H atoms = 2 + 2 = 4

H atoms

• on the LHS = 4
• on the RHS = 2

Thus, adding coefficient "2" in front of $H_{2}O$ on the RHS

$2NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + 2H_{2} O_{(l)}$

Now in RHS,

• Na atoms = 2
• O atoms = 4 + 2 = 6
• H atoms = 4

The no. atoms of each element in the LHS as well as the RHS is equal.

Step 4 :-

Thus, the balanced equation is :-

$2NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + 2H_{2} O_{(l)}$

-   -    -    -    -   -    -    -     -    -    -    -    -    -    -   -   -   -    -   -   -   -   -   -   -   -   -

$\mathbb{THE \: BALANCED \: REACTION :-}$

$2NaOH_{(g)}+H_{2}SO_{4(g)} \overset{}{\longrightarrow}Na_{2}SO_{4(g)} + 2H_{2} O_{(l)}$