Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C3H18 with 4.000 moles of O2. 2 C3H18 + 25 02 → 16 CO2 + 18 H2o 20.00% 25.00% 50.00% 12.50%

94.1g Cr are reacted in the lab and 138g Cr2O3 are weighed. Calculate the percent yield. Pertaining to 4Cr + 3O2 > 2Cr2O3

When 15.0 g of NH3 reacts, the actual yield of N2 is 6.50 g. What is the percent yield?                            4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l) Group of answer choices 30.8% 21.1% 20.6% 31.1 %

If exactly 4.089 moles of propanol (C3H8O)  is combusted in the presence of excess oxygen what is the mass of O2 used in the reaction? The molecular weights of all reactants and products are: C3H8O: 60.09 g/mol O2: 31.998 g/mol H2O: 18.016 g/mol; CO2 : 44.01 g/mol. Enter your answer in the units of g, do not enter the unit itself.

Each step in the following process has a yield of 80.0%.CH4+4Cl2⟶CCl4+4HClCCl4+2HF⟶CCl2F2+2HClThe CCl4 formed in the first step is used as a reactant in the second step. If 2.00 mol CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

Part A Give the theoretical yield, in moles, of CO, from the reaction of 4.00 moles of C3H18 with 4.00 moles of O2. 2 C3H18 + 25 O2 → 16 CO2 + 18 H20

Given the reaction: 2C,H, + 70, → 4C0, + 6H,0 What is the total number of moles of CO, produced by the complete combustion of 5.0 moles of C,H,? 2.0mol 1.0mol 5.0mol O 10.mol

Consider the reaction 3X + 2Y → 5C + 4D How many moles of C can be synthesized from 37.0 moles of Y and excess X?

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows:  Cu2S(s) + O2(G) = 2 Cu (s) + SO2(g)  If the reaction of 0.530 kg of copper(I) sulfide with excess oxygen produces 0.290 kg of copper metal, what is the percent yield?  137% 39.9% 54.7% 274% 68.5%

30) What is the percent yield for the reaction below? Only 28.16 g of CO2 are observed to have formed even though stoichiometry would predict the production of 225.3 g of CO2 since 4.000 X100 moles of C3H18 and 8.000 moles of O2 were used. 2 C8H18+ 25 O2 → 16 CO2 + 18 H2O A) 12.50% B) 25.00% C) 50.00% 20.00%