Question

determine the amount of heat, in kj, given off when 211 g of no₂ is produced according to the following equation:
2 no(g) + o₂(g) → 2 no₂(g)
δh = -114.6 kj/mol
report your answer to the nearest whole number, and be sure to include the correct sign.

Explanation:

Step1: Calculate the molar mass of \(NO_2\)

The molar mass of \(N\) is \(14.01\ g/mol\) and of \(O\) is \(16.00\ g/mol\). For \(NO_2\), \(M=(14.01 + 2\times16.00)\ g/mol=46.01\ g/mol\)

Step2: Calculate the number of moles of \(NO_2\)

Using the formula \(n=\frac{m}{M}\), where \(m = 211\ g\) and \(M = 46.01\ g/mol\). So \(n=\frac{211\ g}{46.01\ g/mol}\approx4.586\ mol\)

Step3: Calculate the heat given off

From the reaction \(2NO(g)+O_2(g)\to2NO_2(g)\), \(\Delta H=- 114.6\ kJ/mol\) (per \(2\) moles of \(NO_2\)). For \(n = 4.586\) moles of \(NO_2\), the heat \(q\) is calculated as follows.
Since for \(2\) moles of \(NO_2\), \(\Delta H=-114.6\ kJ\), for \(1\) mole of \(NO_2\), \(\Delta H=\frac{-114.6}{2}=-57.3\ kJ/mol\)
\(q=n\times\Delta H\) (per mole of \(NO_2\)). Substitute \(n = 4.586\ mol\) and \(\Delta H=-57.3\ kJ/mol\)
\(q=4.586\times(- 57.3)\ kJ\approx - 263\ kJ\)

Answer:

\(-263\ kJ\)