Question

Consider the following thermochemical equation:

Answer 1

Answer:

$m_w=5.3248\times 10^{23}\ kg$

Explanation:

According to the given reaction,

heat released by thecombustion of 2 molecules of Methanol, $\Delta H=1277000\ J$

we know that molecular mass of Methanol, $m=32\ g.mol^{-1}$

∴12 gram of methanol = $\frac{3}{8}\ moles$

we know 1 mole = $6.023\times 10^{23}\ molecules$

so,

$\rm \frac{3}{8} \ moles=2.258\times 10^{23}\ molecules$

Heat from the combustion of $2.258\times 10^{23}\ molecules$:

$Q=1277000\times \frac{2.258\times 10^{23}}{2}$

$Q=1442132.0625\times 10^{23}\ J$

Now the  mass of water that can be heated from 23.5°C to 88.2°C :

$Q=m_w.c_w.\Delta T$

$1442132.0625\times 10^{23}=m_w.4186\times (88.2-23.5)$

$m_w=5.3248\times 10^{23}\ kg$