How many g of KOH are required to completely | vedclass

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

Question

(D) The balanced chemical equation for the neutralization reaction is:$HNO_3 + KOH 	o KNO_3 + H_2O$First,calculate the number of moles of $HNO_3$:Mol

Vedclass · Accepted Answer

$11.2$

Vedclass · Answer

(D) The balanced chemical equation for the neutralization reaction is:$HNO_3 + KOH 	o KNO_3 + H_2O$First,calculate the number of moles of $HNO_3$:Molar mass of $HNO_3 = 1 + 14 + (3 	imes 16) = 63 \, g/mol$Moles of $HNO_3 = \frac{12.6 \, g}{63 \, g/mol} = 0.2 \, mol$According to the stoichiometry of the reaction,$1 \, mol$ of $HNO_3$ reacts with $1 \, mol$ of $KOH$.Therefore,$0.2 \, mol$ of $HNO_3$ requires $0.2 \, mol$ of $KOH$.Now,calculate the mass of $KOH$:Molar mass of $KOH = 39 + 16 + 1 = 56 \, g/mol$Mass of $KOH = 0.2 \, mol 	imes 56 \, g/mol = 11.2 \, g$.

How many $g$ of $KOH$ are required to completely neutralize $12.6 \, g$ of $HNO_3$?

Similar Questions

How many $g$ of a dibasic acid (Mol. wt. $= 200$) should be present in $100 \ mL$ of its aqueous solution to give decinormal strength?

The phosphate of a metal has the formula $MHPO_4$. The formula of its chloride would be

How much of $NaOH$ (in $g$) is required to neutralize $1500 \ cm^3$ of $0.1 \ N \ HCl$? (Given: $Na = 23, O = 16, H = 1$)

The right option for the mass of $CO_2$ produced by heating $20 \ g$ of $20 \%$ pure limestone is $........ \ g$ (Atomic mass of $Ca = 40$) $[CaCO_3 \stackrel{1200 \ K}{\longrightarrow} CaO + CO_2]$

$0.5 \ M$ of $H_2SO_4$ is diluted from $1 \ L$ to $10 \ L$,the normality of the resulting solution is ......... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module