x mg of pure HCl was used to make an aqueous | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The balanced chemical equation for the titration is:$Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_2O_{(\ell)}$Calculate the millimol

Vedclass · Accepted Answer

$1825$

Vedclass · Answer

(B) The balanced chemical equation for the titration is:$Ba(OH)_{2(aq)} + 2HCl_{(aq)} ightarrow BaCl_{2(aq)} + 2H_2O_{(\ell)}$Calculate the millimoles of $Ba(OH)_2$ used:$n_{Ba(OH)_2} = M 	imes V_{(mL)} = 0.1 \ M 	imes 25.0 \ mL = 2.5 \ mmol$From the stoichiometry, $1 \ mol$ of $Ba(OH)_2$ reacts with $2 \ mol$ of $HCl$:$n_{HCl} = 2 	imes n_{Ba(OH)_2} = 2 	imes 2.5 \ mmol = 5.0 \ mmol$Calculate the mass of $HCl$ in $mg$:$Mass = n 	imes Molar \ mass = 5.0 \ mmol 	imes 36.5 \ mg/mmol = 182.5 \ mg$Given $x \ mg = 182.5 \ mg$, we need to express this as $x 	imes 10^{-1}$:$182.5 = 1825 	imes 10^{-1}$Therefore, $x = 1825$.

Similar Questions

$4.0 \, g$ of caustic soda is dissolved in $100 \, mL$ of solution. The normality of the solution is: (in $, N$)

What is the value of percent atom economy if the formula weight of the product is $46 \ u$ and the sum of the formula weight of all reactants is $92 \ u$ (in $\%$)?

For the given series of reactions:
$NH_{3(g)} + O_{2(g)} \rightarrow NO_{(g)} + H_2O_{(l)}$
$NO_{(g)} + O_{2(g)} \rightarrow NO_{2(g)}$
To obtain the maximum mass of $NO_2$ from a given mass of a mixture of $NH_3$ and $O_2$,the ratio of the mass of $NH_3$ to $O_2$ should be:

$A$ metal chloride contains $55.0 \%$ of chlorine by weight. $100 \ mL$ of vapours of the metal chloride at $STP$ weigh $0.57 \ g$. The molecular formula of the metal chloride is $...$. (Given: Atomic mass of chlorine is $35.5 \ u$)

The mass of $CO_2$ that will be obtained by heating $10 \ kg$ of $80 \%$ pure limestone $(CaCO_3)$ is ............... $kg$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module