Question

Name Extra Credit 3. (a) Disulfide, SCl, is produced by the reaction of sulfur with chlorine gas. How many grams of SC may be produced by the reaction of 15.0 g of S with 15.0g of Ca? Step 1: Balance the chemical equation: Se 1Cl2) 45,Cl) Step 2: Determine how many grams of S2Cl2 would be produced from 15.0 g Se. Start with grams of Se, use the molar mass of Se to get moles of Se. 16158mel ast as 0.0lo motes &g tThen, convert to moles of S2Cla by using the molar ratio of SCl; to Sa from the coefficients of the balanced chemical reaction] 0.a4 vol S21z 0 06mol8 4 01 SC1 1pl S [From the moles of S Cla, use the molar mass of S2CI2 to get grams of S2Cl2- 3a.41gSzClz 135.044 5.c1 0-a4 Nol S2C Step 3: Determine how many grams of S2Cl2 would be produced from 15.0 g of Cl2. [Similar to Step 2, but starting from 15.0 g of Clal 1350 4 g/mol S12 4s2CIr X 4 cIr Imel S2c14 70.900 a8-519rams Sacla Step 4: Determine which is the limiting reagent and report how many grams of S2Cl2 are actually produced. Cla Is the limiting reaint (b) In the part 3(a), how many grams of unused reactant will remain once the other reactant is consumed? Step 1: Determine how many grams of the excess reagent was used to react with the limiting reactant. Step 2: Subtract the grams used in the reaction from the starting grams to determine what was left of the excess reagent after the limiting reagent was consumed.

Answer 1

3.ay step-1 Balanced chemical equation, Sg (5)'t sultur thog Yelz (9) Chlorine 150 g 4 S₂ Cl₂ (9) Disulfide Step-2 molar mass of Sall = 135.036 g/mee molar mass of so = 256.52 glmal Moles of so reacts, 15 J Sex I mole se (According = 0.0 58 47 more se 2565234 9 holes of Salla produced. 0.0 58 47 mole sex 4mole S. u. "=0.2338 mole salle to stoichiometric rato) ~ Cerams of S₂ Cl₂ produced from 15g of so 0.2338 mole so c₂ x 135.036 g 6₂ch. I more s, il Imole of Step 3 -= 31.5849xor » molar mass moles of cla of cl-70.906 g/mol reacts, S₂U, 15 g 2 + 1 molech 10.906 9 u, a 20. 2115 mole ce - moles of all 0.2 115 mole , - berams of produced 4 WouSUR umore uz 0.2115 mole so che all prodaced from 15g of a .-135.036g selle = 128.566 g s lx 136 2 Imole sitų - 2008 0.2115 mole S2U

step-y To find limiting reactant, According to stoichiometric ratio, 15 g S8 & 1 molese 256 52 moles& - 0.05 84 molese reaction 1 0 .05847 mile 15 g cx Imolese = 0.05847 70.906 g 0.2115 molex 0.05287 I realtion umore el = 0.05287 0.15847 ch the I be will be the limiting realtant encess reactant and Se another opproalh : To calmlate trinimum nimum grams grams at moles at of one reactant (ay A) needed to completely consumed a second reactanth say B . If the stoichiometric available amount of realtant 'A' enceeds what is needed then it is eness neactant, if not, then the reactant B will be limiting reactant. Starting with so we calmlate minimum grams of c2 needed to completely consumed 15 g sg. 15 g sex I moe so & 4 mole e 70.9069 - 256 52 hole so Imole sgt Imolect = 16-5848 15.0g U2 available to us in than the 16.158 g uz needed I will be the the limiting reactant. Tout to us is ele less la will be the the limitin so and so cle will be grams of completely consumed. S. Uz actually produced 1 X 4 x 1 ч, - что 4 moles, lh 70.902. goch more che & 135.036 g selle Imoufler = 28.568 g 28.568 g Soch

from the above calculation we cansaf Stepay a will be the encess reactant so grams of 'se' needed to compume el completely 15g U2 x 1 1 molech Imole Sq mole sh 256:52 g S8 70.902 " I mole sg = 13.5673 g So Step-2 But 150 g of så available to us. . the Gerams of eness reactant left after limiting nealtant was consumed = ( 15 - 13.5773) g 1.427