Question

in the following equations, determine which reactant is the limiting reactant and which reactant is in excess. the amounts used are given below.

koh + hno₃→ kno₃ + h₂o

13.2 g koh and 12.0 g hno₃ react.

the limiting reactant is:

○ hno₃
○ h₂o
○ kno₃
○ koh

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Explanation:

Step1: Calculate moles of reactants

Molar mass of \(KOH = 39 + 16+1=56\space g/mol\). Moles of \(KOH=\frac{13.2\space g}{56\space g/mol}\approx0.236\space mol\)
Molar mass of \(HNO_3 = 1 + 14+3\times16 = 63\space g/mol\). Moles of \(HNO_3=\frac{12.0\space g}{63\space g/mol}\approx0.190\space mol\)

Step2: Use mole ratio from balanced equation

From \(KOH + HNO_3
ightarrow KNO_3 + H_2O\), mole ratio of \(KOH:HNO_3 = 1:1\)
Since \(0.190\space mol\) (moles of \(HNO_3\)) \(<0.236\space mol\) (moles of \(KOH\))

Answer:

A. \(HNO_3\)