When 6.00 g оf CS2 (MM = 76.13 g/mоl) gаs reаct with 10.0 g оf Cl2 (MM = 70.90 g/mol) gаs in the following reaction: CS2(g) + 3 Cl2(g) → CCl4(l) + S2Cl2(l) What is the limiting reactant? How many grams of CCl4 (153.81 g/mol) are produced? How many grams of the excess reactant are left? If 6.50 g of CCl4 are produced, what is the percent yield of the reaction? Must show your work on scratch paper to receive credit.