In the following reaction, oxygen is the excess reactant. SiCl4 + O2 → SiO2 + Cl2 The table shows an experimental record for the above reaction. Experimental Record Trial Starting Amount of SiCl4 Starting Amount of O2 Actual Yield of SiO2 1 120 g 240 g 38.2 g 2 75 g 50 g 25.2 g Calculate the percentage yield for SiO2 for Trial 1. Also, determine the leftover reactant for the trial. Show your work. Based on the percentage yield in Trial 2, explain what ratio of reactants is more efficient for the given reaction.

For the following reaction, 6.37 grams of phosphoric acid are mixed with excess potassium hydroxide. The reaction yields 11.3 grams of potassium phosphate. potassium hydroxide (aq) + phosphoric acid (aq) →potassium phosphate (aq) + water (I) What is the theoretical yield of potassium phosphate ? grams What is the percent yield of potassium phosphate ? % Submit Answer Retry Entire Group 4 more group attempts remaining

Question 86 of 97 According to the balanced reaction below, calculate the moles of NH; gas that form when 4.2 mol of N2H. liquid completely reacts: 3 N2H.(0)→ 4 NH3(g) + N2(g) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 4.2 28.02 5.6 17.04 16.8 3.2 32.06 6.022 x 10 mol NH3 mol N2H4 g N2 g NH3 g N2H4 mol N2 MacBook Air

esc Calculate the quantity of O, would be required to generate 13.0 mol of NO in the reaction below assuming the reaction has only 72.4% yield. 2 NO (g) + O₂(g) → 2 NO₂ (g) ! 1 S Q A O @ 2 W S CE #3 E D $ 4 R % 5 F LL T MacBook Pro ^ 6 G Y & 7 FEB 21 H zoom U 8 W -

K $ 4 R For the following reaction, 4.16 grams of hydrogen gas are allowed to react with 45.2 grams of iodine hydrogen (g) + iodine (s)→ hydrogen iodide (g) What is the maximum amount of hydrogen iodide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? Submit Answer % 5 [Review Topics] [References] Use the References to access important values if needed for this question. Show Hint T Retry Entire Group 4 more group attempts remaining Cengage Learning Cengage Technical Support < 6 Y d 7 U 8 I < ( 9 0 O P grams grams * Previous 4) [

For the following chemical reactions, how do I identify the number of moles of the compound in the third column that would be produced from the number of moles of compound in the second column assuming that all of the other reactants in the equation in the first column are in excess.                   Reaction   Moles initially     Moles produced N2 + 3H2 → 2 NH3 0.027 moles N2 a.______moles NH3 5 C + 2 SO2 →CS2 + 4 CO 3.80 moles C b._______moles CO 5 C + 2 SO2 →CS2 + 4 CO 6.7 moles C c._______moles CS2

Determine the total mass of the product that could be formed from the balanced reaction shown below if one starts with 4.20 g Al (26.98 g/mol) and 17.40 g iodine, I₂ (MW 253.80 g/mol). 2Al + 3I₂ → 2AlI₃     Set up the table below that represents 100% yield with the given reaction conditions..

MISSED THIS? Watch KCV: Limiting Reactant, Theoretical Yield, and Percent Yield, IWE: Limiting Reactant and Theoretical Yield; Read Section 4.4. You can click on the Review link to access the section in your eText. Iron(II) sulfide reacts with hydrochloric acid according to the reaction FeS (s) + 2HCl(aq)→FeCl₂ (aq) + H₂S(s) A reaction mixture initially contains 0.212 mol FeS and 0.636 mol HCl. Part A Once the reaction has reached completion, what amount (in moles) of the excess reactant is left? Express the answer in moles to three significant figures. n = Submit V ΑΣΦ Request Answer ? mol

How many grams of N2 are needed to completely (stoichiometrically) react with  222 g of Br2 ?  N2   +   3 Br2   →   2 NBr3   +   44.0 kJ              The molar mass of Br2 is _____________ g and the molar mass of N2 is  ___________ g                                ____________ mol Br2                                                              (2.22 x 102 g Br2)(––––––––––––––––––––––––––––––)  =  __________ mol Br2                               _______________ g Br2                                                                                                                       ______________ mol N2 __________________ mol Br2 (–––––––––––––––––––––––––) =  _______________ mol N2                                                     __________________ mol Br2                                                       ___________________ g N2 _______________ mol N2 (–––––––––––––––––––––––––) =  _____________ g N2                                                    ________________ mol N2

In the following reaction, oxygen is the excess reactant.SiCl4 + O2 → SiO2 + Cl2The table shows an experimental record for the above reaction. Experimental Record Trial Starting Amount of SiCl4 Starting Amount of O2 Actual Yield of SiO2 1 150 g 200 g 49.2 g 2 75 g 50 g 25.2 g Calculate the percentage yield for SiO2 for Trial 1. Also, determine the leftover reactant for the trial. Show your work. Based on the percentage yield in Trial 2, explain what ratio of reactants is more efficient for the given reaction.                     Source                  StylesNormal