Respuesta :
[tex]\boxed{2172.1{\text{ kg}}}[/tex] of sodium carbonate must be added to neutralize [tex]2.01 \times {10^3}{\text{ kg}}[/tex] 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:
[tex]{\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}}}[/tex]
According to stoichiometry of reaction, one mole of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] reacts with one mole of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] to produce one mole of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex], one mole of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] and one mole of [tex]{\text{C}}{{\text{O}}_{\text{2}}}[/tex]. So stoichiometric ratio between [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] and [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] is 1:1.
The formula to calculate number of moles of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] is as follows:
[tex]{\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}}}}}[/tex] …… (1)
Substitute [tex]2.01 \times {10^3}{\text{ kg}}[/tex] for mass of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] and 98.079 g/mol for molar mass of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] in equation (1).
[tex]\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}[/tex]
Since one mole of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] reacts with one mole of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex], 20493.7 moles of [tex]{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] reacts with 20493.7 moles of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex].
The formula to calculate mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] is as follows:
[tex]{\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)[/tex] …… (2)
Substitute 20493.7 mol for moles of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] and 105.9888 g/mol for molar mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] in equation (2).
[tex]\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}[/tex]
Learn more:
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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.
The amount of sodium carbonate that would be used to neutralize 2.01 x [tex]10^3[/tex] kg of sulfuric acid would be 2.172 x [tex]10^3[/tex] kg
The reaction between sodium carbonate and sulfuric acid is represented by the following equation:
[tex]Na_2CO_3(aq)+H_2SO_4(aq)---->Na_2SO_4(aq)+CO_2(g)+H_2O(l)[/tex]
The mole ratio of the reactants is 1:1. Meaning that 1 mole of Na2CO3 is required for every 1 mole of H2SO4 spill.
Mole of 2.01 x [tex]10^3[/tex] kg of H2SO4 = mass in grams/molar mass
= 2.01 x 10^3 x 1000/98.079
= 20,493.6837 moles
Thus, 20,493.6837 moles of Na2CO3 would also be needed.
Mass of 20,493.6837 moles Na2CO3 = mole x molar mass
= 20,493.6837 x 105.9888
= 2,172,100.939 g
= 2,172 kg
= 2.172 x [tex]10^3[/tex] kg
More on calculating moles can be found here: https://brainly.com/question/12513822
