Question

plz show work

LIONS <Chapter 7 Yield Calculations Simulation - Limiting Reagent and Theoretical Yield with Masses www. ③ 7013 Extended applications for the law of conservation of mass Although it serves as the principle by which we establish the stoichiometric relationships between reactants and products, the law of conservation of mass can be used more rechy to determine values of mass when seemingly critical information is missing such as the identity of the substance and, with it the moms). The mass of newly discovered meshare been deduced (albeit sometimes through complex calculations) based on the law of conservation of mass, but simpler determinations can be made, such as the identity of an own hydrocarbon through combustion. We will consider how mass can be used to determine an unknown number of molecules for a reaction with an unknown hydrocarbon Part D As with any combustion reaction, the products of combusting a hydrocarbon fuel (C,H,) with oxygen (0) are carbon dioxide (CO) and water (HO) A mass of 15.35 g for an unknown fuel was combusted in a reaction vessel containing an unknown amount of oxygen. At the end of the reaction, there still remained 15.31 g of the fuel well as 0.0654 g of water and 0.1198 g of carbon dioxide. The oxygen was completely consumed during the reaction How many molecules of oxygen gas were initially present in the reaction vesser? Express the number of molecules to three significant digits. View Available Hint(s) VALD RO? molecules amount of O, = Submit Next? Provide Feedback

Answer 1

Date: - According to Combustion Reaction عالعدد يقطعك Og should be (ar + 2 / 2 No of Moles for because in product side there are хохуля іn 2x oxygen in co, & 7 in Ho a Hence on reactant side there should be (2x + y ) Oxygen he o As we have Og i we divide (2x +y by a to make equal no. of moles - on both sides So balanced Chemical equation is Getty + (2x+ 912) O x + (2) H20 /

Date: Now, according 0.06 54g of is produced to H₂O Reaction & 0.1198 given g of ca We know, no. of moles (n) .. Mass Molar Mass Co. HO (Molor C. Mas). ") 44.0 g 18.0 g h(coo) : 0.1198 0.0027 moles. 44 n(H₂O) 0.0654 18 0.0036 moles x 0.0027 * 0.0036 Alsor h. No of Molecules Avagar's No. (NA) 2x + 4/2 0.0045 moles n (for 0,) No. of molecules: = 0.0045 X 0.027 X 10 6.0 23 Xlo" molecules of ₂