6. Calculate the standard enthalpy change for the reaction 2CaHi (l) + 1 702(g)

Question

6. Calculate the standard enthalpy change for the reaction 2CaHi (l) + 1 702(g) ? 16C0(g) + 181 1201). Given: 2CsH,s(l) + 250tg) ?16COdg) + 18H20(1) ?l®--11,020 kJ/mol ?? 10-.-566.0 kJ/mol 2CO(g) + Odg)-+2CO2(g) A) -10.450 kJ/mol B) -6,492 kJ/mol C) 6,492 kJ/mol D) 10,450 kJ/mol E) 15,550 kJ/mol 7. The combustion of pentane produces heat according to the equation CsH 12(1) + 80;(g) ? 5C0;(g) + 61120(l) ?10,--3,510 kJ/mol How many grams of CO2 are produced per 2.50 x 103 kJ of heat released? A) 0.0809 g B) 3.56 g C) 31.3 g D) 157 g E) 309 g 8. Given the thermochemical equation 2S0:(g)+ 0x(g)->2S0s(g). AH PN-198 kJ/mol, how much heat is evolved when 600. g of SO2 is burned? A) 5.46 x 102 kJ B) 928 kJ C) 1.85 x 103 kJ D) 3.71x 103 kJ E) 59,400 kJ Determine the heat given off to the surroundings when 9.0 g of aluminum reacts according to the equation 2A1 + Fed) ? AIO, + 2Fe, AHO.--849 kJ/mol. A) 1.4 x10H B) 2.8 x 102 kJ C) 5.6 x 102 kJ D) 2.5 x 103 kJ E) 7.6 x 103 kJ 9.

Explanation / Answer

6)

The reaction we need to get is:
2C8H18(l) + 17O2(g) ------> 16CO(g) + 18H2O(l)

For that we have the reaction 1 which is: 2C8H18(l) + 25O2(g) ------> 16CO2(g) + 18H2O(l)

And the reaction 2: 2CO(g) + O2(g) ------->2CO2(g)

All we need to do is invert the second reaction, (The value of H° is also inverted in sign) and then multiply that by 8, so we can annul the CO2 values and substract the O2:

2C8H18(l) + 25O2(g) ------> 16CO2(g) + 18H2O(l)
(2CO2(g) -----------> 2CO(g) + O2(g))x8

2C8H18(l) + 25O2(g) ------> 16CO2(g) + 18H2O(l)   H° = -11020 kJ/mol
16CO2(g) -----------> 16CO(g) + 8O2(g)   H° = 8*566 = 4528 kJ/mol

2C8H18(l) + 17O2(g) ------> 16CO(g) + 18H2O(l)  

H° = -11020 + 4528

= -6492 kJ/mol

7)

The given equation is

C5H12(l) + 8O2(g) ? 5CO2(g) + 6H2O(l)

H = - 3510 KJ

Calculate the moles of CO2 required for 2500 kJ of heat released

When 3510 kJ of heat release = 5 mol of CO2 produced

For 2500 kJ of heat released = 5 x 2500/3510 = mol of CO2 produced

=  3.56 mol CO2 produced

Convert the moles of CO2 into grams = moles x molecular weight

= 3.56 mol/ 44 g/mol

= 157 grams

Option D is the correct answer

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