**Concord High School Chemistry**

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**Practice Problem #2:**

Rb_{3}AsO_{4} + BaCr_{2}O_{7} –>

**Answer to Practice Problem #2:**

**1.** **Identify the type of reaction.**

Double-replacement reaction.

Rb^{+} is a cation, so it will switch with Ba^{2+} (also a cation).

Rb_{3}AsO_{4} + BaCr_{2}O_{7} –>

**2.** **Write out the products of the chemical reaction.**

Switch the **cations**, since this is single-replacement:

Rb_{3}AsO_{4} + BaCr_{2}O_{7} –> Rb Cr_{2}O_{7} + Ba AsO_{4}

Then write the formulas of the products based on their charges.

Rb = +1, Cr_{2}O_{7} = -2, so the formula is Rb_{2}Cr_{2}O_{7}.

Ba = +2, AsO_{4}= -3, so the formula is Ba_{3}(AsO_{4})_{2}.

Rb_{3}AsO_{4} + BaCr_{2}O_{7} –> Rb_{2}Cr_{2}O_{7 } + Ba_{3}(AsO_{4})_{2}

**3.** **Balance the equation.**

Rb_{3}AsO_{4} + BaCr_{2}O_{7} –> Rb_{2}Cr_{2}O_{7 } + Ba_{3}(AsO_{4})_{2}

Balance this equation by putting the coefficients in front of the compounds.

Last, double check that all numbers of atoms of each element is equal.

**Answer to Problem:** **2Rb _{3}AsO_{4} + 3BaCr_{2}O_{7} –> 3Rb_{2}Cr_{2}O_{7 } + Ba_{3}(AsO_{4})_{2}**

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