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H2S; H: 2 (1.01) S: 32.06 34.089/mol 02:2(16) = 329/mol H20: H: 2(1,01) Section 11.3 1) Consider the unbalanced equation: 2 H2S(s) + 3 02(g) →2 SO2(g) +2 H20 () %3D a. How many grams of sulfur dioxide could be produced from 53.0 g H2S? (34.08) = 68.16 53.09H,S. 128.129 sor 53.0gA;S. 128.129 soz 6 16 11.bg S0z produced %3D 18.02 9 (mo) (14.06)-128.12 S02: S: 32.06 0: 2(16) b. If 148.0 g of SO2 are released how many grams of O2 were there to start with? OL 3(32-0) =96 148.04 S02 9690t goz 64.06g lmol %3D 12812 178.12 c. If 178.0 g of H2S are mixed with 157.0 g O2 which reactant is limiting and how much (in grams) sulfur dioxide can be produced? 178.0 g HeS, Iol eS 34.08 HyS Oz is the limiting reactant because it wll run out Zmol 902 5,223 mol S0 z %3D 2mol H25 first. 137.0gor. Inator Znol sz 3.27 mol S0z 3metoz 157.0g 0z. 3matoz 128.129 soz SOz. 314.29 a g Soz 2mol Soz d. If only 198.7 g of sulfur dioxide were produced in the lab from part c. what was the percent yield?

H2S; H: 2 (1.01) S: 32.06 34.089/mol 02:2(16) = 329/mol H20: H: 2(1,01) Section 11.3 1) Consider the unbalanced equation: 2 H2S(s) + 3 02(g) →2 SO2(g) +2 H20 () %3D a. How many grams of sulfur dioxide could be produced from 53.0 g H2S? (34.08) = 68.16 53.09H,S. 128.129 sor 53.0gA;S. 128.129 soz 6 16 11.bg S0z produced %3D 18.02 9 (mo) (14.06)-128.12 S02: S: 32.06 0: 2(16) b. If 148.0 g of SO2 are released how many grams of O2 were there to start with? OL 3(32-0) =96 148.04 S02 9690t goz 64.06g lmol %3D 12812 178.12 c. If 178.0 g of H2S are mixed with 157.0 g O2 which reactant is limiting and how much (in grams) sulfur dioxide can be produced? 178.0 g HeS, Iol eS 34.08 HyS Oz is the limiting reactant because it wll run out Zmol 902 5,223 mol S0 z %3D 2mol H25 first. 137.0gor. Inator Znol sz 3.27 mol S0z 3metoz 157.0g 0z. 3matoz 128.129 soz SOz. 314.29 a g Soz 2mol Soz d. If only 198.7 g of sulfur dioxide were produced in the lab from part c. what was the percent yield?

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Can someone please check to see if I did this correctly? I feel like o light be doing that last part of c wrong which would also make d wrong. Thank you!
Section 11.3 1) Consider the unbalanced equation: HeS; Hi 2 (1.01) S: 32.06 2 H2S(s) + 3 Ö2(g) → 2 SO2(g) 2 H2O (1) 34.089/mol 02:2(16) = 32 a. How many grams of sulfur dioxide could be produced from 53.0 g H2S? H20: H: 2(1,01) (34.08) = 68.16 53.09H,S. 128.129 sor 6 16 11.bg S0z produced 18.02 9(mo) So2: S: 32.06 0: 2016) (4.00)=128.12 b. If 148.0 g of SO2 are released how many grams of O2 were there to start with? 3(32.0) =96 148.09 Sozx 969 0z "goz 64.06g lmol 128.12 c. If 178.0 g of H2S are mixed with 157.0 g O2 which reactant is limiting and how much (in grams) sulfur dioxide can be produced? 178.0 q HeS, Imol HeS 34.08 HyS Oz is the limiting reactant because it wl run out Zmol 302 5,223 mol 90z 2mol HzS first. Znol soz 3.27 mol Soz 32goz 3metoz 157.0g 0z. 3 motoz. 128.129 soz 2mol Soz 314.29 a g Soz d. If only 198.7 g of sulfur dioxide were produced in the lab from part c. what was the percent yield? % yeild = experimental value 198,79 302 So2 X100= 63.276 calcu lated Value 기니. 29q S0z 1
Transcribed Image Text:Section 11.3 1) Consider the unbalanced equation: HeS; Hi 2 (1.01) S: 32.06 2 H2S(s) + 3 Ö2(g) → 2 SO2(g) 2 H2O (1) 34.089/mol 02:2(16) = 32 a. How many grams of sulfur dioxide could be produced from 53.0 g H2S? H20: H: 2(1,01) (34.08) = 68.16 53.09H,S. 128.129 sor 6 16 11.bg S0z produced 18.02 9(mo) So2: S: 32.06 0: 2016) (4.00)=128.12 b. If 148.0 g of SO2 are released how many grams of O2 were there to start with? 3(32.0) =96 148.09 Sozx 969 0z "goz 64.06g lmol 128.12 c. If 178.0 g of H2S are mixed with 157.0 g O2 which reactant is limiting and how much (in grams) sulfur dioxide can be produced? 178.0 q HeS, Imol HeS 34.08 HyS Oz is the limiting reactant because it wl run out Zmol 302 5,223 mol 90z 2mol HzS first. Znol soz 3.27 mol Soz 32goz 3metoz 157.0g 0z. 3 motoz. 128.129 soz 2mol Soz 314.29 a g Soz d. If only 198.7 g of sulfur dioxide were produced in the lab from part c. what was the percent yield? % yeild = experimental value 198,79 302 So2 X100= 63.276 calcu lated Value 기니. 29q S0z 1
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