Questions: Making Activity Series Predictions Use the activity series below to predict the products of each of the following reactions. You do not need to balance the equations. Li > K > Ba > Sr > Ca > Na > Mg > AI > Mn > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > H > Sb > Bi > Cu > Ag > Pd > Hg > Pt > Au Na2 CO3 + K → ? no reaction K2 CO3 + Na NaK + CO3

Making Activity Series Predictions

Use the activity series below to predict the products of each of the following reactions. You do not need to balance the equations.

Li > K > Ba > Sr > Ca > Na > Mg > AI > Mn > Zn > Cr > Fe > Cd >
Co > Ni > Sn > Pb > H > Sb > Bi > Cu > Ag > Pd > Hg > Pt > Au

Na2 CO3 + K → ?

no reaction
K2 CO3 + Na
NaK + CO3
Transcript text: Making Activity Series Predictions Use the activity series below to predict the products of each of the following reactions. You do not need to balance the equations. \[ \begin{array}{l} \mathrm{Li}>\mathrm{K}>\mathrm{Ba}>\mathrm{Sr}>\mathrm{Ca}>\mathrm{Na}>\mathrm{Mg}>\mathrm{AI}>\mathrm{Mn}>\mathrm{Zn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Cd}> \\ \mathrm{Co}>\mathrm{Ni}>\mathrm{Sn}>\mathrm{Pb}>\mathrm{H}>\mathrm{Sb}>\mathrm{Bi}>\mathrm{Cu}>\mathrm{Ag}>\mathrm{Pd}>\mathrm{Hg}>\mathrm{Pt}>\mathrm{Au} \end{array} \] \[ \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{K} \rightarrow \text { ? } \] no reaction $\mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{Na}$ $\mathrm{NaK}+\mathrm{CO}_{3}$
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Solution

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Solution Steps

Step 1: Identify Reactants and Products

We are given the reactants \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) and \(\mathrm{K}\). We need to predict the products of this reaction using the activity series.

Step 2: Compare Reactivity

According to the activity series, potassium (K) is more reactive than sodium (Na). This means that potassium can displace sodium in a compound.

Step 3: Predict the Products

Since potassium is more reactive, it will replace sodium in \(\mathrm{Na}_{2}\mathrm{CO}_{3}\), forming \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and releasing sodium.

Final Answer

\[ \boxed{\mathrm{K}_{2}\mathrm{CO}_{3} + \mathrm{Na}} \]

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