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asambeis [7]
3 years ago
14

Please select the word from the list that best fits the definitions

Chemistry
1 answer:
Mamont248 [21]3 years ago
7 0
Erm i’m afraid we need the list
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odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add
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Answer : The mass of sodium carbonate added to neutralize must be, 6.54\times 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}

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Molar mass of H_2SO_4 = 98 g/mole

Mass of H_2SO_4 = 6.05\times 10^3kg=6.05\times 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=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol

The moles of H_2SO_4 is, 6.17\times 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\times 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\times \text{ Molar mass of }Na_2CO_3

Molar mass of Na_2CO_3 = 106 g/mole

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

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

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