Asked Oct 10, 2019

How many liters of Cl2(g) at STP can be made from NaCl by oxidation with 24.0 grams of K2Cr2O7 in an acidic solution after chromium is reduced to Cr3+

6NaCl + K2Cr2O7 + 7H2SO4 = 3Cl2 + Cr2(SO4)3 + 3Na2SO4 + K2SO4+ 7H2


Expert Answer

Step 1


Grams of K2Cr2O7 = 24.0 g

Moles of K2Cr2O7 can be calculated as:


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294 g/mol Molar mass of K2Cr2O7= (2x39) + (2x52)+ (7x16) 24 g Mass 0.0816 mol Moles 294 g/mol Molar mass

Step 2

The balanced chemical reaction is given as:

6 NaCl + K2Cr2O7 + 7 H2SO4 → 3 Cl2 + Cr2(SO4)3 + 3 Na2SO4 + K2SO4+ 7 H2

From the reaction, it is evident that:


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1 mol K2Cr2O7 produces 3 mol Cl2 3 mol Cl2 :. 0.0816 mol K2Cr2O7 produces x 0.0816 mol K2Cr2O7 1 mol K2Cr207 - 0.2448 mol Cl2

Step 3

The ideal gas equatio...


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PV nRT Where, P pressure V = volume n number of moles R gas constant = 0.0821 L-atm/mol-K T temperature


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