Question #f77cd

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Question #f77cd

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Answer 1 · 2017-09-04T20:12:39

The concentration of the phosphoric acid is $1.23 \times 10^{-3} l {mol\cdotdm}^{-3}$ $1.23 × 10^"-3" color(white)(l)"mol·dm"^"-3"$ .

Step 1. Write the balanced equation

$M_{r} : m l l 310.18 m m m m m m m m m m m m m m m m 98.00$ $M_text(r):color(white)(mll)310.18color(white)(mmmmmmmmmmmmmmmm)98.00$
$m m m {Ca}_{3} {({PO}_{4})}_{2} + {3HNO}_{3} \to {6Ca(NO}_{3})_{2} + {2H}_{3} {PO}_{4}$ $color(white)(mmm)"Ca"_3("PO"_4)_2 + "3HNO"_3 → "6Ca(NO"_3)_2 + "2H"_3"PO"_4$

Step 2. Calculate the moles of ${Ca}_{3} {({PO}_{4})}_{2}$ $"Ca"_3("PO"_4)_2$

$"Moles of Ca"_3("PO"_4)_2 = 7.26 color(red)(cancel(color(black)("g Ca"_3("PO"_4)_2))) × (1 "mol Ca"_3("PO"_4)_2)/(310.18 color(red)(cancel(color(black)("g Ca"_3("PO"_4)_2)))) = "0.023 41 mol Ca"_3("PO"_4)_2$

Step 3. Calculate the moles of $"H"_3"PO"_4$

$"Moles of H"_3"PO"_4 = "0.023 41" color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("2 mol H"_3"PO"_4)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "0.046 81 mol H"_3 "PO"_4$

Step 4. Calculate the molarity of the $"H"_3 "PO"_4$

$"Molarity" = "moles"/"cubic decimetres" = "0.046 81 mol"/"38.0 dm"^3 = 1.23 × 10^"-3" color(white)(l)"mol·dm"^"-3"$

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