Given the following Information: CaO(s) + H_2O(l) rightarrow Ca(OH)_3(g) Delta H

Question

Given the following Information: CaO(s) + H_2O(l) rightarrow Ca(OH)_3(g) Delta H degree x n = -64 8 k j/mol How many grams of CaO must react in order to liberate 525 Kj of heat? 625 kj of heat? 6.92 g 56.18 455 g 606 g 3.40.10 times 8 Given the thermochemical equation 2SO_2(g) +O_2(g) rightarrow 2SO_2(g) Delta H degree min = -196 kJ How much heat is evolved when 600 g of SO_2 is burned? 5.46 times 10^-2 Kj. 928 kJ 1.85 times 10^3 kj 59.400 kJ 3.71 times 10^3 kJ Pentaborone B5H9(s) burns vigorously in O_2 to give B2O3(g) and H20(l) according to equation; 2B_3H_9 + 12O_2rightarrow 5B_2O_3 + 9H_2. Calculate the Delta H for the combustion of 1 mol of B5H9 given the following enthalpies of formation. Delta H degree [B_2O_3(g)] = -1, 273.5 K j/mol. Delta H degree [B_5H_3(g)] = 73.2 K j/mol. Delta H degree [H_3(g)] = -285.8 K j/mol. Delta H degree [O_3(g)] = 0 K j/mol. -12735 K j. -4543 K j. -19170 K j. -9086 K j. -8448 K j. The combustion of butane produces heat according to the equation. 2 C_4 H_12(g) + 13 O_2(g) rightarrow 8 CO_2(g) + 10 H_2O(l) Delta H degree _max = - 5, 314 K j. 1.20 g. 139 g. 0.0413 g. 69.7 g. 97.8 g.

Explanation / Answer

Q15

For

Q = 525 kJ heat we need

HRxn = -64.8 kJ/mol of CaO

so...

mol of CaO needed:

Q = -n*HRxn

n = -Q/HRn = -525/(-64.89)= 8.0906 mol of CaO required

CaO MW = 56.0774 G/mol

mass = mol*MW = (8.0906)(56.0774)

mass of CaO = 453.69 g required

choose C

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