# 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+ 7H2O

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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

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Step 1

Given,

Grams of K2Cr2O7 = 24.0 g

Moles of K2Cr2O7 can be calculated as: help_outlineImage Transcriptionclose294 g/mol Molar mass of K2Cr2O7= (2x39) + (2x52)+ (7x16) 24 g Mass 0.0816 mol Moles 294 g/mol Molar mass fullscreen
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: help_outlineImage Transcriptionclose1 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 fullscreen
Step 3

The ideal gas equatio... help_outlineImage TranscriptionclosePV nRT Where, P pressure V = volume n number of moles R gas constant = 0.0821 L-atm/mol-K T temperature fullscreen

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