2:15 AM Sun Apr 14 Question 8 of 8 In this reaction: Mg (s) + 12 (s) → Mgl2 (s) If 1.84 moles of Mg react with 3.56 moles of 12, and 1.76 moles of Mgl2 form, what is the percent yield? Tap here or pull up for additional resources % 1 2 3 ☑ 4 5| 6 C 7 8 9 +/- 0 ×10 5% Submit
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- 1:39 PM Sun Apr 14 Question 8 of 8 In this reaction: Mg (s) + 12 (s) → Mgl2 (s) If 1.84 moles of Mg react with 3.56 moles of 12, and 1.76 moles of Mgl2 form, what is the percent yield? Tap here or pull up for additional resources % " 97% ■ 1 2 3 ☑ 4 5| 6 C 7 8 9 +/- 0 ×10 Submit< 16:10 1 4 7 +/- Question 1 of 4 If 2.35 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g) LO 00 2 3 6 8 9 5 mol O Tap here or pull up for additional resources 68 Submit XU x 100Molar mass pf Ca (IO3)2 =40+2 (126+3 (16)) =40+2(126+48) =40+2 (174) = 40+348 =388g/mol Mass of Ca(IO3)2 produced = moles x molars mass of Ca (IO3)2 = 0.01875 x 388g= 7.275g (c)What is the percent yield if the actual yield of Ca(IO,), is 6.72 g ? %3D www ww
- 8. Calculate the percentage yield of t-butyl chloride if we obtained 9.3g product from 16.7 mL of t-butyl alcohol and excess HCI (t-butyl alcohol: M= 74 g/mol, d= 0.78 g/ml; t-butyl chloride: M=92.5g/mol). [Chose the nearest answer). (a) 73% (b) 81% (c) 57% (d) 41% (e) 65%O Macmillan Learning ...pdf Consider the reaction. 2 Pb(s) + O₂(g) →→→ 2 PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 385.4 g of lead(II) oxide. Calculate the percent yield of the reaction. percent yield: F PDF x10 TOOLS Lab 4 BIOL22010L....pdf I Attem 37°F Cloudy ^ % 12:C 2/21,Balance the given chemical equation and then complete the reaction table based on the starting conditions represented in the atomic-scale picture. If you determine that a species should have a coefficient of 1, you must indicate this with the "1" label. Do not leave the coefficient space blank for any species in the balanced equation. = H2 1 = N2 H2(g) N2(g) NH3(g) Initial Chenn 14 of 15 Next > < Prev asus 2) 3. 4.
- Given the balanced reaction, initial GRAMS of each reactant and product, and the molar masses of each species complete the remainder of the change and final rows, in GRAMS. NOTE: Report all masses with ONE DECIMAL PLACE and be sure to include + and - signs with numerical values with NO SPACES. 2 C2H6 (g) + 7 O2 (g) ----> 4 CO2 (g) + 6 H2O (g) C2H6 (MM = 30.07 g/mol) O2 (MM = 32.00 g/mol) CO2 (MM = 44.01 g/mol) H2O (MM = 18.02 g/mol) Initial Grams 10.0 10.0 0.0 0.0 Change Grams Final GramsIdentify the excess reagent. Balanced Chemical equation: 2 Na + 3 N, 2 NaN, Amount Reagent Max. Amount of product 2.5 moles NaN, 2.5 Moles Na 6 Moles N, 4 moles NaN,E Effects of Ho X My Home Brake Assignmen: TakeCovaler Activity doplocator assignment-take Al X X Submit Answer E ContentSen What is the theoretical yield of phosphoric acid ? What is the percent yield of phosphoric acid ? O Show Hint X OWLv2 | Onlin For the following reaction, 4.11 grams of tetraphosphorus decaoxide are mixed with excess perchloric acid (HCIO4). The reaction yields 4.77 grams of phosphoric acid. perchloric acid (HCIO4) (aq) + tetraphosphorus decaoxide (s)-phosphoric acid (aq) + dichlorine heptaoxide (1) M Retry Entire Group 2 more group attempts remaining X grams % G cengage ans X + Previous Email Instructor Feb 5 Save and Exit 9:43 8 G I
- Question 1 of 4 Submit In the reaction below, 7.0 mol of NO and 5.0 mol of O, are reacted together. The reaction generates 3.0 mol of NO2. What is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g) | % 1 3 4 6. C 7 9. +/- x 10 0 Tap here or pull up for additional resources LO 00les/23530384?wrap%3D1 Hwk 17 MOLE.docx OLE.docx 17 MOLE.docx (13.3 KB) ZC Hwk 17 MOLE/ MOLE STOICHIOMETRY 1) First balance the equation The combustion of a sample of butane, C,H10 (lighter fluid), produced 5.75 mol of water. C,H10 + O2 -------> CO,+ H,O a) How many moles of butane burned? b) How much oxygen was used up in moles?app.101edu.co G Paused Question 5 of 12 Submit Determine the number of grams of C«H10 that are required to completely react to produce 8.70 mol of CO2 according to the following combustion reaction: 2 CAH10(g) + 13 O2(g) – 8 CO2(g) + 10 H2O(g) moleH10 g CAH10 8.70 moteo. STARTING AMOUNT 8 moteo. 2.18 moleH1o 2 mol C4H10 g C4H10 8.70 mol CO2 x 8 mol CO2 2.18 mol C4H10 ADD FACTOR DELETE ANSWER RESET *( ) 10 126 13 2.18 1 8 6.022 x 1023 18.02 0.0374 4.35 8.70 44.01 58.14 208.00 g O2 g CAH10 g H2O mol H20 g CO2 mol CO2 mol O2 mol C4H10