Answer to Question #131805 in General Chemistry for fsecret

Question

. Superphosphate, or sometimes called triple phosphate, is a commonly used water-soluble fertilizer. It is produced by the reaction: 𝐶𝑎3 (𝑃𝑂4 )2 + 𝐻2𝑆𝑂4 → 𝐶𝑎(𝐻2𝑃𝑂4 )2 + 𝐶𝑎𝑆𝑂4 If we mix 200.0 g of 𝐶𝑎3 (𝑃𝑂4 )2 with 133.5 g of 𝐻2𝑆𝑂4, how much 𝐶𝑎(𝐻2𝑃𝑂4 )2 (in grams) can be produced?

Answer 1

The Chemical equation is,

$Ca_3(PO_4)_2\ +H_2SO_4\ \to Ca(H_2PO_4)_2\ +CaSO_4\ where,\newline Ca_3(PO_4)_2=200g\ and,\newline H_2SO_4=133.5g$

A balanced chemical equation of the reaction is

$Ca_3(PO_4)_2\ +2H_2SO_4\ \to\ Ca(H_2PO_4)_2\ +\ 2CaSO_4$

Calculate the number of moles of each reagent.

$moles=\dfrac{mass}{molar\ mass}$

$Ca_3(PO_4)_2\ mols\ =\dfrac{200}{310.18}\ =0.6448\ mol$

$H_2SO_4\ mols=\dfrac{133.5}{98.079}\ = 1.361\ mol$

Express moles of each reagent in terms moles of the product

$Ca(H_2PO_4)_2\ moles=0.6448*\dfrac{1 molCa(H_2PO_4)_2}{1\ molCa_3(PO_4)_2}$

$=0.6448mol$

$Ca(H_2PO_4)_2\ moles=1.361*\dfrac{1\ molCa(H_2PO_4)_2}{2moles H_2SO_4}$

$=0.6805mol$

Since this is limiting reagent problem, the reagent that produces a smaller number of moles of the product is used in calculating the mass of the product.In this case $Ca_3(PO_4)_2$ has the smaller number of moles .

$Mass=no.moles\ of\ Ca_3(PO_4)_2*\ molar \ massCa(H_2PO_4)_2$

$=0.6448*234.1\newline =150.9g$