Question

question 5
calculate the energy (in kj) during the reaction involving 9 moles of reactant b, if the heat of reaction is -955 kj/mol given the
balanced reaction: 2 a + 3 b → c + 3 d. show your answer to 3 significant figures and do not include units. pay attention to the sign
of your answer.
add your answer

Explanation:

Step1: Determine moles of reaction

From the balanced reaction \(2A + 3B
ightarrow C + 3D\), 3 moles of B correspond to 1 mole of reaction. So for 9 moles of B, moles of reaction \(n=\frac{9}{3}=3\) moles.

Step2: Calculate energy change

The heat of reaction \(\Delta H=- 955\space kJ/mol\) (per mole of reaction). For \(n = 3\) moles of reaction, energy \(E=\Delta H\times n=-955\times3=-2865\), rounded to 3 significant figures is \(-2.87\times10^{3}\) or \(-2870\) (but in 3 sig figs, \(-2.87\times10^{3}\) is \(-2870\) when written as such, but more accurately \(-2.87\times10^{3}\) which is \(-2870\) with 3 sig figs). Wait, \(955\times3 = 2865\), so with 3 sig figs, it's \(2.87\times10^{3}\), and with the sign, \(-2.87\times10^{3}\) or \(-2870\) (but \(2865\) rounded to 3 sig figs is \(2870\) (since the fourth digit is 5, we round up the third: \(2865
ightarrow2870\) when considering 3 sig figs? Wait, \(2865\): first three digits 2,8,6. The next digit is 5, so round 6 up to 7, so 2870. So \(-2870\) (but in scientific notation, \(-2.87\times10^{3}\)).

Answer:

\(-2.87\times10^{3}\) (or \(-2870\))