1) MgI 2 +Br 2 (yields) MgBr 2 +I 2 a) Which is the excess reactant when 560 g o
ID: 677178 • Letter: 1
Question
1) MgI2 +Br2 (yields) MgBr2 +I2 a) Which is the excess reactant when 560 g of MgI2and 360 g of Br2 react, and what mass remains? b) What mass of I2 is formed in the sameprocess? 2) CS2 + 3 O2 (yields) 2 SO2+ CO2 a) if 1.60 mol of CS2 burns with 5.60 mol ofO2 , how many moles of the excess reactant will still bepresent when the reaction is over? 1) MgI2 +Br2 (yields) MgBr2 +I2 a) Which is the excess reactant when 560 g of MgI2and 360 g of Br2 react, and what mass remains? b) What mass of I2 is formed in the sameprocess? 2) CS2 + 3 O2 (yields) 2 SO2+ CO2 a) if 1.60 mol of CS2 burns with 5.60 mol ofO2 , how many moles of the excess reactant will still bepresent when the reaction is over? a) Which is the excess reactant when 560 g of MgI2and 360 g of Br2 react, and what mass remains? b) What mass of I2 is formed in the sameprocess? 2) CS2 + 3 O2 (yields) 2 SO2+ CO2 a) if 1.60 mol of CS2 burns with 5.60 mol ofO2 , how many moles of the excess reactant will still bepresent when the reaction is over?Explanation / Answer
find the limiting reactent which ever one leads to least amount of moles. so 2.01 mols of MgI2 and 2.25 mols Br2 since there is less MgI2 than Br2 that means for a br2 is excess so2.25 moles - 2.01 moles = .24 moles of excess br2 which then equals38.35 grams of br2 using stoichiometry its 1 to 1 so there will form 2.01 mol ofI2 so times molar mass its 510 g of I2 now for 2 same idea for ever cs2 u need 3 mols of oxygen so lets compare o2 to 3x cs2 so u would need 4.8 mol for ever 1.6 mol of cs2 that means 5.6-4.8 = .8 mole of excess oxygen
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