Question

6. What is the limiting reagent in the following reaction: moar geHu PAY amal Chy al CD cD2 32-al 0z al co,- Co .07u6 mal alr alysi mass Lin 1of CD 347mei CH,(g)+ 20,(g)-co()+ 2H,0(g) 23.50 734 mal leal given only 12.0 g CH, and 23.5 g O,? What is the theoretical yield of CO,? ion (.Su7mat)(uu.01 . 2g cor 7. If the above experiment gave 12.5 g of CO2(g), what is the percent yield?

Answer 1

mass ef CH4 - 12g тn оlоцесии, mass moru mass imel -0.75 mol moles ef coly = 0.ts mol mass of o2 = 23.5g moles of o2= 23.5g - 32g|mat +34 mel. mores of og = 0.234 mes. considering Reaction - ( Balanced) Hy (g) + 202 (9) cog (g) + 2 Hgolgy I I 0.75 m al 0.734one. steps to f the limiting Reactant- 1) find the arailable mores- Coy = 0. Is mel. q = 0.734 mel.

Divide the moles Stoichiometric to their corresponding cofficients (S.(s) eing CH4) 0.75ml = o.fs mol. 2 0.7 34 nel - 0.367 malo 1" S.C The least value of mele is the limiting Reagent (IR) 0.367 nel roots mol. og z limiting Reagent (LR) nalue of limiting Regent | LoR) = 0.36f mol. moles of product formed a sic of that yllR) prodent moes of coa formed - 11 x 0.367) mol. moles of log prodened - 003 67 mal. mass ef cog produced: 0.367 molt erg map = 16.15g on

mass ef log produced = 16.15g Theorefical geild of log 16.15g Experimental mass of coga 12.5 g Actual yeild of log = 12.5g - % Yeild & - - Actual yeild - Plooy Theoretical yeild % yeild a 12.58 xinol Thisg Ć 77.4.). %0 yeild = ff.4%
O2: limiting reagent

Theoretical yield=16.15g

% yield=77.4%