Question

Consider the reaction 7H2SO4+Na2Cr2O7+6NaCl=Cr2(SO4)3+3Cl2+4Na2SO4+7H2O
Assume the reaction is quantitative. If a mixture of 16.7 g H2SO4, 3.46g Na2Cr2O7, and 4.23g NaCl is caused to react,
a) Which will be the limiting reactant? (NaCl)
b) How many grams of Cl2 can be formed? (2.56g Cl)
c) How many grams of which reactants will be left over? (Sodium Chromate .0524g H2SO4 8.43 g) I believe answer is in ( ), but need a detailed explanation of how to get it. Thank you so much!







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Answer 1

a)
Moles of NaCl = 4.23g/58.44g/mol = 0.0723mol
Moles of Na2Cr2O7 = 3.46g/261.98g/mol = 0.0132mol
Moles of H2SO4 = 16.7g/ 98.09g/mol = 0.1702 mol

6molNaCl for 1 mol Na2Cr2O7 = 0.0723/6 = 0.01205 mol ----> limiting

7 mol H2SO4 for 1 mol Na2Cr2O7 = 0.1702/7 = 0.0243142 mol

b) (0.0723mol)NaCl * (3 mol Cl2/6 mol NaCl) * (70.9g/1 mol Cl2) = 2.56 g Cl2

c) 0.0723mol NaCl* 7 mol H2SO4/6 mol NaCl = 0.08436 mol H2SO4 = 0.1702-0.08436= * 98.09g/ 1 mol = 8.42 g H2SO4

try the last one yourself...