Question

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kgkg of sulfuric acid solution?

Answer 1

Answer : The mass of sodium carbonate added to neutralize must be,
`6.54* 10^3kg`

Explanation :

First we have to calculate the moles of
`H_2SO_4`.

`\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}`

Given:

Molar mass of
`H_2SO_4` = 98 g/mole

Mass of
`H_2SO_4` =
`6.05* 10^3kg=6.05* 10^6g`

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

`\text{Moles of }H_2SO_4=(6.05* 10^6g)/(98g/mol)=6.17* 10^4mol`

The moles of
`H_2SO_4` is,
`6.17* 10^4mol`

Now we have to calculate the moles of
`Na_2CO_3`

The balanced neutralization reaction is:

`Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3`

From the balanced chemical reaction we conclude that,

As, 1 mole of
`H_2SO_4` neutralizes 1 mole of
`Na_2CO_3`

So,
`6.17* 10^4mol` of
`H_2SO_4` neutralizes

Now we have to calculate the mass of
`Na_2CO_3`

`\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3* \text{ Molar mass of }Na_2CO_3`

Molar mass of
`Na_2CO_3` = 106 g/mole

`\text{ Mass of }Na_2CO_3=(6.17* 10^4mol)* (106g/mole)=6.54* 10^6g=6.54* 10^3kg`

Thus, the mass of sodium carbonate added to neutralize must be,
`6.54* 10^3kg`