Question

50 g of CaCO3 is allowed to react with 70 g of H3PO4 Calculate :
(i) amount of Ca3(PO4)2 formed
(ii) amount of unreacted reagent​

Answer 1

Answer:

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Explanation:

Given:

CaCO3 =50g

H3PO4 =70g

then the amount formed I'd calculated by this formula= total mass of compounds /2 ., then

50+70/2= 60g

then the mass of Ca³(PO4)² is 60g and the unreacted reagent is calculated by this formula=

Max. Compound -Min. Compound = 70-50

20g is unreacted reagent

Answer 2

Given:

Mass of CaCO3 = 50 gm

Mass of H3PO4 = 70 gm

To Find:

(i) amount of Ca3(PO4)2 formed

(ii) amount of unreacted reagent​

Calculation:

- The balanced chemical equation is:

3 CaCO3 + 2 H3PO4 → Ca3(PO4)2 + 3 CO2 + 3 H2O

- 3 mole of CaCO3 react with 2 moles of H3PO4 to form 1 mole of Ca3(PO4)2.

⇒ 300 gm of CaCO3 reacts with 196 gm of H3PO4.

- And the Mass of Ca3(PO4)2 formed = 310 gm

(i) 50 gm of CaCO3 will reacts with 32.67 gm of H3PO4.

- Mass of Ca3(PO4)2 formed = (310/300) × 50

⇒ M = 51.67 gm

(ii) Here, CaCO3 is the limiting reagent. So, H3PO4 is left unreacted.

Mass of unreacted H3PO4 = 70 - 32.67

⇒ Unreacted H3PO4 = 37.33 gm

- So, 51.67 gm of Ca3(PO4)2 is formed and 37.33 gm of H3PO4 is left unreacted.