The following reacations are occurring in solvay process used for the manufacture of sodium carbonate. (B) + H 2 O → (C) (E) + CO 2 (excess) → (F) Given atomic masses: Na = 23, K = 39, Ca = 40, Mg

Question

The following reacations are occurring in solvay process used for the manufacture of sodium carbonate.

(B) $+ H_{2} O \to$ (C)

(E) $+ CO_{2}$ (excess) → (F)

Given atomic masses: Na = 23, K = 39, Ca = 40, Mg = 24, Fe = 56, Cu = 63.5, O = 16, S = 32, H = 1, C = 12, Cl = 35.5, F = 19, Zn = 65, N_A = $6.022 \times 1 0^{23}$ .

The number of molecules of (F) obtained on reaction of $1/2$ mole of (E) with $1/4$ mole of CO₂ is $Q \times 1 0^{21}$ .

What is the value of $Q$ ?

TestHub · Accepted Answer

G = NaHCO 3 ​ , D = NH 3 ​ , E = NH 4 ​ OH , F = NH 4 ​ HCO 3 ​ The reaction between (E) and C O 2 ​ to form (F) occurs in a 1:1 molar ratio. Since C O 2 ​ has fewer moles ( 0.25 moles) than (E) ( 0.5 moles), C O 2 ​ is the limiting reagent. Therefore, exactly 0.25 moles of (F) are produced. Total molecules = 0.25 × 6.022 × 1 0 23 = 1.5055 × 1 0 23 . Converting this to the power of 1 0 21 gives 150.55 × 1 0 21 , so Q = 150.55 .

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