How many g of KOH are required to completely | vedclass

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(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

$11.2$

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(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 $HNO_3$?

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