How many milliliters of 0.736 M H3PO4 are required to react with 238 mL of 0.160 M Ba(OH)2 if the...

Question:

How many milliliters of {eq}0.736 \, \mathrm{M \; H_3PO_4} {/eq} are required to react with {eq}238 \, \mathrm{mL} {/eq} of {eq}0.160 \, \mathrm{M \; Ba(OH)_2} {/eq} if the products are barium phosphate and water?

Reaction stoichiometry calculations:

The reaction between phosphoric acid and barium hydroxide to produce barium phosphate and water is a double displacement reaction. The stoichiometric coefficients obtained by balancing the reaction equation can be used to determine the amounts of reactants needed.

Answer and Explanation:

Step 1: Write a balanced reaction equation.

{eq}\rm 2H_3PO_4+3Ba(OH)_2\rightarrow Ba_3(PO_4)_2+6H_2O {/eq}

Step 2: Determine the moles of {eq}\rm Ba(OH)_2 {/eq} in 238 mL of 0.160 M solution.

Since by definition of molarity, 1000 mL = 0.160 moles the moles of {eq}\rm Ba(OH)_2 {/eq} in 238 mL of 0.160 M solution can be obtained as follows.

{eq}\rm moles \ of \ Ba(OH)_2=\frac{0.160 \ mol}{1000 \ mL}\times 238 \ mL {/eq}

{eq}\rm moles \ of \ Ba(OH)_2=0.0381 \ moles {/eq}

Step 3: Use the stoichiometric ratios to determine the moles of {eq}\rm H_3PO_4 {/eq} that will react with 0.0381 moles of {eq}\rm Ba(OH)_2 {/eq}.

The mole ratio of {eq}\rm H_3PO_4 \ : \ Ba(OH)_2 {/eq} is 2 : 3. The moles of {eq}\rm H_3PO_4 {/eq} can be obtained as follows.

{eq}\rm moles \ of \ H_3PO_4=\frac{2 \ mol \ of \ H_3PO_4}{3 \ mol \ of \ Ba(OH)_2}\times 0.0381 \ mol \ of \ Ba(OH)_2 {/eq}

{eq}\rm moles \ of \ H_3PO_4=0.0254 \ moles {/eq}

Step 4: Determine the volume of 0.736 M solution that contains 0.0254 moles of {eq}\rm H_3PO_4 {/eq}.

By definition of molarity 0.736 moles are in 1000 mL of the solution. The volume of the solution that contains 0.0254 moles of {eq}\rm H_3PO_4 {/eq} can be obtained as follows.

{eq}\rm Volume =\frac{1000 \ mL}{0.736 \ mol}\times 0.0254 \ mol {/eq}

{eq}\rm Volume =\textbf{34.5 mL} {/eq}


Learn more about this topic:

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Mass-to-Mass Stoichiometric Calculations

from Chemistry 101: General Chemistry

Chapter 9 / Lesson 3
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