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s2008m [1.1K]
3 years ago
14

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?
Chemistry
1 answer:
gayaneshka [121]3 years ago
8 0
THe balanced chemical reaction would be:
Na2CO3 + H2SO4 = Na2SO4 + H2O + CO2

We first convert the mass of sulfuric acid solution to moles by the molar mass. Then, we relate H2SO4 with Na2CO3 from the reaction. We do as follows:

<span>6.05×10^3 kg H2SO4 ( 1 kmol / 98.08 kg) ( 1 kmol Na2CO3 / 1 kmol H2SO4 ) ( 105.99 kg / kmol ) = 6537.92 kg Na2CO3 needed</span>
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If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut
Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl  = moles of OH⁻ from Ca(OH)2  so :

0.204 * 42.8 / 1000  => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles (  since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass =  0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2


Hope that helps!

6 0
3 years ago
A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b
solmaris [256]
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>

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