一题Chemistry问题求解（Heat of displacement）

kangaroo99

When 4.8 g og magnesium is added to 50cm^3 of 2.0 mol/dm^3 copper cloride solution, the temperature of the mixture increases by 4.0 °C. What is the heat of reaction in this experiment?
[Specific heat capacity of solution = 4.2 J g^-1 °C^-1 ; relative atomic mass of Mg = 24]


A -0.42 J mol^-1
B -0.84 J mol^-1
C -4.20 kJ mol^-1
D -8.4 kJ mol^-1

书上的答案是C，可是我算到却是D。

Mg + CuCl2 → MgCl2 + Cu


我心想no. of mole of Mg = 4.8/24 = 0.2 mol
然后no. Of mole of MgCl2 =  [50(2)]/1000 = 0.1 mol

我心想跟着equation来比较就拿到Mg是in excess，所以0.1 mol of Mg will react with 0.1 mol of Cu^2+to produce 0.1 mol of MgCl2 and 0.1mol of Cu


然后就用mc(theta)来算heat，拿到840J， 然后再找heat of displacement
0.1 mol → 840J
1 mol → 8400 J
所以Heat of reaction是-8.4 kJ mol^-1

罗宾航

我也是算到D，为什么楼主你的算法那么复杂。。。我的很简单罢了。。。

kangaroo99

你怎样算分享下

化学我自修所以都是跟着参考书的方式