Question

10g of CaCO3 are treated with a solution of Hc, 1.12L of CO2 are produced at STP. Find out the number of moles of HOI​

Answer 1

Answer:

(Sec A + Cos A) (Sec A - Cos A)

We know that,

(a - b) (a + b) = a² - b²

Here,

a = Sec A

b = Cos A

Substituting,

Sec² A - Cos ² A

We know that,

1 + Tan² A = Sec² A

1 - Sin² A = Cos² A

Substituting,

1 + Tan² A - (1 - Sin² A)

1 + Tan² A - 1 + Sin² A

Tan² A + Sin² A

Therefore,

\boxed{\purple{\textsf{(Sec A + Cos A) (Sec A - Cos A) = Tan² A + Sin² A}}}

(Sec A + Cos A) (Sec A - Cos A) = Tan² A + Sin² A

Answer 2

Answer:

CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O

So 1- mole of CaCO3 will react with 2- moles of HCl.

10- g of CaCO3=0.1-mole (100m.mole)

250- ml of 2N HCl= 0.5- mole (500- m.mole)

So 100- m.moles of CaCO3 will react with 200- m.moles of HCl.

So (500–200)= 300- m.moles of HCl will remain unreacted in the 250- ml of solution.

So Normality of HCl = 300 x 4/1000-N

Normality of HCl = 1.2 N