An impure sample of solid na₂co₃ is allowed to react

Ответ:

Kay6718

02.01.2022

Chemistry

Thus percentage purity of sodium carbonate is 60.83%.

Explanation:

$Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O$

Volume of the HCl solution ,V = 15.55 mL= 0.01555 L

Concentration of HCl solution ,C= 0.1755 M

Moles of HCl in 0.1755 M solution = n

$Molarity=\frac{n}{V(L)}$

$n=C\time V=0.1755 M\times 0.01555 L = 2.729\times 10^{-3} mol$

According o reaction 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then n moles of HCl will react with:

$\frac{1}{2}\times n=\frac{1}{2}\times 2.729\times 10^{-3} mol=1.3645\times 10^{-3} mol$ of sodium carbonate.

Mass of $1.364\times 10^{-3} mol$ of sodium carbonate:

$1.364\times 10^{-3} mol\times 106 g/mol=0.144584 g$

Mass of the sample given= 0.2377 grams

Mass used up in reaction = 0.144585 g

percentage purity= $\frac{0.144585}{0.2337}\times 100=60.83\%$

Thus percentage purity of sodium carbonate is 60.83%.

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Ответ:

Shadow0202

22.08.2021

Chemistry

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. Each electron is negatively charged. The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons.

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An impure sample of solid na₂co₃ is allowed to react with 0.1755 m hcl.
$na_2co_3 + 2hcl \longrightarrow 2nacl + co_2 + h_2o$
a 0.2337-g sample of sodium carbonate requires 15.55 ml of hcl solution. what is the purity of the sodium carbonate?

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An impure sample of solid na₂co₃ is allowed to react with 0.1755 m hcl. a 0.2337-g sample of sodium carbonate requires 15.55 ml of hcl solution. what is the purity of the sodium carbonate?

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An impure sample of solid na₂co₃ is allowed to react with 0.1755 m hcl.
$na_2co_3 + 2hcl \longrightarrow 2nacl + co_2 + h_2o$
a 0.2337-g sample of sodium carbonate requires 15.55 ml of hcl solution. what is the purity of the sodium carbonate?