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

no, i was listening to water by schoolboy q while and i saw this question

explenation:

water isn't wet. wetness is a description of our experience of water; what happens to us when we come into contact with water in such a way that it impinges on our state of being. we, or our possessions, 'get wet'. a less impinging sense experience of water is that it is cold or warm, while visual experience tells us that it is green or blue or muddy or fast-flowing. we learn by experience that a sensation of wetness is associated with water: 'there must be a leak/i must have sat in something.'

Answer from: Quest

Higher temperature = molecules move around faster. and vice versa.