20 mL of 0.1 M HCl is added to 30 mL of 0. | vedclass

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

Question

(B) The reaction is: $HCl + NaOH ightarrow NaCl + H_2O$. Number of moles of $HCl = 0.1 \ M 	imes 0.02 \ L = 0.002 \ mol$. Number of moles of $NaOH

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(B) The reaction is: $HCl + NaOH ightarrow NaCl + H_2O$. Number of moles of $HCl = 0.1 \ M 	imes 0.02 \ L = 0.002 \ mol$. Number of moles of $NaOH = 0.1 \ M 	imes 0.03 \ L = 0.003 \ mol$. Since $HCl$ is the limiting reagent,it will be completely consumed. Number of moles of $NaOH$ remaining after neutralization $= 0.003 \ mol - 0.002 \ mol = 0.001 \ mol$. Total final volume $= 20 \ mL + 30 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. Molarity of the final solution $= \frac{0.001 \ mol}{0.1 \ L} = 0.01 \ M$.

$20 \ mL$ of $0.1 \ M$ $HCl$ is added to $30 \ mL$ of $0.1 \ M$ $NaOH$. To this solution,an extra $50 \ mL$ of water was added. What is the molarity of the final solution formed (in $M$)?

Similar Questions

Calculate the weight of $FeO$ produced from $6.7 \ g$ of $VO$ and $4.8 \ g$ of $Fe_2O_3$ using the reaction: $2VO + 3Fe_2O_3 \rightarrow 6FeO + V_2O_5$ $(V = 51, Fe = 56)$ (in $g$)

$5 \ mL$ of $0.1 \ M \ Pb(NO_{3})_{2}$ is mixed with $10 \ mL$ of $0.02 \ M \ KI$. The amount of $PbI_{2}$ precipitated will be about

$3O_2 + 2N_2 \to 2N_2O_3$
$9 \ mol$ of $O_2$ and $14 \ mol$ of $N_2$ are allowed to react. When $3 \ mol$ of $O_2$ remains unreacted,how many moles of $N_2O_3$ would have been produced?

Assign $A, B, C, D$ from the given type of reaction.
$NaBr + AgNO_3 \longrightarrow AgBr \downarrow + NaNO_3$

When $30 \ mL$ of $H_2$ reacts with $20 \ mL$ of $O_2$ to form water,what is leftover when the reaction ends?

Vedclass Test Series

Exam Paper Generator

Online Exam Module