Question

50g of caco3 is allowed to react with 73.5 of h3po4. calculate -:1) amount of ca3(po4)2

Answer 1

Done your question downside

Answer 2

Answer:

An amount of 75.66 g of  $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)$ is found.

Explanation:

Balancing the equations $3 \mathrm{CaCO}_{3}+2 \mathrm{H}_{3} \mathrm{PO}_{4}-\rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

So 210g of $\mathrm{CaCO}_{3}$and 196g of$\mathrm{H}_{3} \mathrm{PO}_{4}$ will react to give 310g of $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$

Now as given in the question 50 g of $\mathrm{CaCO}_{3}$ is taken to let it react with 73.5 g of $\mathrm{H}_{3} \mathrm{PO}_{4}$

Then applying simple unitary method we get  

X=(50+73.5)$\times \frac{310}{210}+196$

On solving the above equation

we get X= 75.66 g

So, the value of $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)$ is 75.66 grams