How much boat it released when 327 g. of oxygen gat u used In the following reac

Question

How much boat it released when 327 g. of oxygen gat u used In the following reaction? 4 Bi_(s) + 3O_2_(g) rightarrow 2 Bi_2O_3_(s) Delta H^+ = -8.55 kJ suppose you mix 20.5 g of water at 66.2 degreee c with 45.4 of water at 35.7 degreee C in an insulated container. What is the final temperature reached after mixing? Use the following thermochemical equations to calculate the Delta H degreee_rxn for the following reaction. Be careful!! only use the necessary reactions in your calculations! 3y_2O_5(s) + 7CO(g) rightarrow 7 CO_2(g) + 2Y_3O_4(s) -deltaH_rxn = ?? 2y_3O_4(s) + CO_2(g) rightarrow 3Y_2O_3(s) + CO(g) Delta H degreee = -380.0 kJ CO(g) + 2 YO_2(s) rightarrow CO_2(g) + Y_2O_3(s) Delta H degreee = +121.8 kJ CO_2(g) + 3Y_2O_5(g) rightarrow CO(g) + Y_3O_4(s) Delta H degreee = +211.2 KJ 2YO_2(s) + CO_2(g) rightarrow Y_2O_5 (s) rightarrow Y_2O_5(s) + CO(g) Delta H degreee = -344.5 kJ

Explanation / Answer

8) moles of O2 = 327 g/32 g/mol = 10.22 mol

3 moles gives off -855 kJ heat

So, 10.22 mol would give heat = -855 x 10.22/3 = 2912.7 kJ

9) Heat lost by hot water = heat gained by cold water

let Tf be the final temperature

Heat q = mCpdT

20.5 x 4.18 x (66.2 - Tf) = 45.4 x 4.18 x (Tf - 35.7)

1357.1 - 20.5Tf = 45.4Tf - 1620.78

2977.88 = 65.9Tf

so final temperature Tf = 45.20 oC

10) Multiply equation 4 with 3 and invert reaction : dH = 3 x 344.5 = 1033.5 kJ

Invert equation 1 : dH = 380 kJ

Multiply equation 2 with 3 : dH = 3 x 121.8 = 365.4 kJ

Add new 1,2 and 4 equations

dHrxn = 1033.5 + 380 + 365.4 = 1778.9 kJ

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