Chemistry, 07.07.2019 11:30 mary603

Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to neutralize 2.01×103 kg of sulfuric acid solution?

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Answer from: jeanieb

$\boxed{2172.1{\text{ kg}}}$ of sodium carbonate must be added to neutralize $2.01 \times {10^3}{\text{ kg}}$ of sulfuric acid solution.

Further Explanation:

Stoichiometry:

Theamountof species present in the reaction is determined with the help of stoichiometryby the relationship between reactants and products. It is used to determine the moles of a chemical species when moles of other chemical species present in the reaction is given.

Balanced chemical reaction between sodium carbonate and sulphuric acid is as follows:

${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} + {{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}} + {\text{C}}{{\text{O}}_{\text{2}}}$

According to stoichiometry of reaction, one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ reacts with one mole of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ to produce one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ , one mole of ${{\text{H}}_{\text{2}}}{\text{O}}$ and one mole of ${\text{C}}{{\text{O}}_{\text{2}}}$ . So stoichiometric ratio between ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is 1:1.

The formula to calculate number of moles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is as follows:

${\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = \dfrac{{{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}}}$ …… (1)

Substitute $2.01 \times {10^3}{\text{ kg}}$ for mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ and 98.079 g/mol for molar mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} &= \left( {\frac{{2.01 \times {{10}^3}{\text{ kg}}}}{{{\text{98}}{\text{.079 g/mol}}}}} \right)\left( {\frac{{{{10}^3}{\text{ g}}}}{{1{\text{ kg}}}}} \right)\\&= 20493.7{\text{ mol}}\\\end{aligned}$

Since one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ reacts with one mole of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ , 20493.7 moles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ reacts with 20493.7 moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ .

The formula to calculate mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is as follows:

${\text{Mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = \left( {{\text{Moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right)\left( {{\text{Molar mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right)$ …… (2)

Substitute 20493.7 mol for moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and 105.9888 g/mol for molar mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ in equation (2).

$\begin{aligned}{\text{Mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} &= \left( {{\text{20493}}{\text{.7 mol}}} \right)\left( {{\text{105}}{\text{.9888 g/mol}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kg}}}}{{1{\text{ g}}}}} \right)\\&= 2172.1{\text{ kg}}\\\end{aligned}$

Learn more:

Calculate the moles of chlorine in 8 moles of carbon tetrachloride: link Calculate the moles of ions in the solution: link

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, Na2CO3, H2SO4, Na2SO4, CO2, H2O, balanced chemical reaction, 1:1, one mole, stoichiometric ratio, 2172.1 kg.

Answer from: Quest

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –

6.022

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,

will have a gram-molecular weight of 352g.

Answer from: Quest

what's the

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Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodiu...

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