Respuesta :
Answer:
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
Explanation:
Mass of sulfuric acid solution = [tex]6.05\times 10^3 kg=6.05\times 10^6 g[/tex]
[tex]1 kg = 10^3 g[/tex]
Percentage mass of sulfuric acid = 95.0%
Mass of sulfuric acid = [tex]\frac{95.0}{100}\times 6.05\times 10^6 g[/tex]
[tex]=5,747,500 g[/tex]
Moles of sulfuric acid = [tex]\frac{5,747,500 g}{98 g/mol}=58,647.96 mol[/tex]
[tex]H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O[/tex]
According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.
Then 58,647.96 moles of sulfuric acisd will be neutralized by :
[tex]\frac{1}{1}\times 58,647.96 mol=58,647.96 mol[/tex] of sodium carbonate
Mass of 58,647.96 moles of sodium carbonate :
[tex]106 g/mol\times 58,647.96 mol=6,216,683.76 g[/tex]
6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg
6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.
We have that for the Question "How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution" it can be said that
- 5748 kg of H2SO4 =6217 kg of Na_2CO_3
From the question we are told
- A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0%H2SO4 by mass and has a density of 1.84 g/mL.
- Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?
Generally the equation for the 95% mass of sulfuric acid is mathematically given as
[tex]95\% \ mass \ of\ sulfuric\ acid= 6.05*1000*0.95[/tex]
[tex]95\% \ mass \ of\ sulfuric\ acid = 5748g[/tex]
Therefore
The resultant equation
[tex]Na_2CO_3+ H_2SO_4 >>>> Na_2SO_4+ H_2O+CO_2[/tex]
Therefore
with the appropriate molar masses we discover that
106 kg of [tex]Na_2CO_3[/tex] needs 98 kg of [tex]H_2SO_4[/tex] for a reaction to take place
Therefore
[tex]5748 kg\ of\ H2SO4 =\frac{5747.5*106}{98}[/tex]
5748 kg of H2SO4 =6217 kg of Na_2CO_3
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