Question

Phosphoric acid can be manufactured according to the following skeleton reactions (unbalanced)
$C{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}+Si{{O}_{2}}+C+{{O}_{2}}+{{H}_{2}}O\to CaSi{{O}_{3}}+C{{O}_{2}}+{{H}_{3}}P{{O}_{4}}$
The equal masses of calcium phosphate (M=310) and silica(M=60) are reacted with excess carbon, oxygen and water to produce $1.96\times {{10}^{3}}$kg phosphoric acid (M=98). What mass (in kg) of calcium phosphate was used, assuming 100% yield?

Answer 1

To solve this question, we need the balanced chemical equation because a balanced equation tells us the exact amount of moles used by the different compounds and the moles of the products formed. The weight will be calculated from the moles of the compounds.

Complete step by step solution:
-Mole is the amount of substance that contains the atoms, molecules or other particles in an entity equal to the atoms present in 12 g of C-12 isotope. It defines the basis of physical chemistry and so its number is called Avogadro's number which is equal to 6.023x${{10}^{23}}$ entities. It is denoted by the symbol ${{N}_{A}}$ .
-The reaction for the process given in the question occurs in multiple steps. These reactions are

& C{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}+3Si{{O}_{2}}\to 3CaSi{{O}_{3}}+{{P}_{2}}{{O}_{5}} \\\ & 2{{P}_{2}}{{O}_{5}}+10C\to {{P}_{4}}+10CO \\\ & 2CO+{{O}_{2}}\to 2C{{O}_{2}} \\\ & {{P}_{4}}+6{{H}_{2}}O+5{{O}_{2}}\to 4{{H}_{3}}P{{O}_{4}} \\\ \end{aligned}$$ -The products formed in the preceding reactions are consumed in the succeeding reactions and so they cancel out with each other after balancing and only the essential compounds remain. The complete balanced chemical equation for the given question can be written as $$C{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}+3Si{{O}_{2}}+C+{{O}_{2}}+3{{H}_{2}}O\to 3CaSi{{O}_{3}}+C{{O}_{2}}+2{{H}_{3}}P{{O}_{4}}$$ -The moles of the atoms and compounds need to be the same throughout the reaction. Mole is written as $$mole=\dfrac{wt.\text{ in grams}}{molecular\text{ wt}\text{.}}$$ -Let the mass of calcium phosphate and silica be m. Their moles will be represented as $\dfrac{m}{310}\text{ and }\dfrac{m}{60}$ respectively. If the moles reacted are x, then the moles remaining of these compounds will be $\dfrac{m}{310}\text{-x and }\dfrac{m}{60}-3x$ respectively. The phosphoric acid formed will be 2x. -Now we are given the weight of the phosphoric acid. From that, we can find the value of x as \begin{aligned} & 2x= \dfrac{1.96 \times {{10}^{3}}}{98} \\\ & \Rightarrow \text{x=10} \\\ \end{aligned} -As the entire reactant was consumed, we can equate the following relation for calcium phosphate $\begin{aligned} & \dfrac{m}{310}\text{-x=0} \\\ & \Rightarrow \dfrac{m}{310}\text{-10=0} \\\ & \Rightarrow m=3100 \\\ & \\\ \end{aligned}$ **Therefore the weight of calcium phosphate in kg is 3100.** **Note:** Mole is not only related to mass but also to volume as any matter that has a certain mass will occupy a certain volume. 1 mole is the amount of molecules present in 22.4 litres of a gas calculated at standard temperature and pressure.