A 10 g effervescent tablet containing sodium | vedclass

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(D) The chemical reactions are:$2NaHCO_3 + H_2C_2O_4 	o Na_2C_2O_4 + 2H_2O + 2CO_2$Let the mass of $NaHCO_3$ be $x \ g$. Then the mass of $H_2C_2O_4$

Vedclass · Accepted Answer

$8.4$

Vedclass · Answer

(D) The chemical reactions are:$2NaHCO_3 + H_2C_2O_4 	o Na_2C_2O_4 + 2H_2O + 2CO_2$Let the mass of $NaHCO_3$ be $x \ g$. Then the mass of $H_2C_2O_4$ is $(10 - x) \ g$.Moles of $NaHCO_3 = \frac{x}{84}$.Moles of $H_2C_2O_4 = \frac{10 - x}{90}$.Since $2 \ mol$ of $NaHCO_3$ reacts with $1 \ mol$ of $H_2C_2O_4$,the limiting reagent determines the $CO_2$ produced.Assuming $NaHCO_3$ is the limiting reagent,$n_{CO_2} = n_{NaHCO_3} = \frac{x}{84}$.Given $n_{CO_2} = \frac{0.25 \ L}{25.0 \ L \ mol^{-1}} = 0.01 \ mol$.$\frac{x}{84} = 0.01 \implies x = 0.84 \ g$.Percentage of $NaHCO_3 = \frac{0.84 \ g}{10 \ g} 	imes 100 = 8.4\%$.

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