Question

Although coal is a complex mixture of substances, its elemental composition can be approximated by the formula /tex.z-dn.net/?f=C_%7B135%7DH_%7B96%7DO_9NS" id="TexFormula1" title="C_{135}H_{96}O_9NS" alt="C_{135}H_{96}O_9NS" align="absmiddle" class="latex-formula">. Using this formula, predict the amount of CO₂ released from the combustion of 1.00 × 10⁶ metric tons of coal (about the annual average for a coal-fired power plant). 1 metric ton = 1 ×10³ kg.
a. 3.66 × 10⁹ kg
b. 3.11 × 10⁹ kg
c. 8.50 × 10⁸ kg
d. 3.11 × 10⁵ kg

Answer 1

Answer:

The correct answer is option b.

Explanation:

$2C_{135}H_{96}O_9NS+313O_2\rightarrow 270CO_2+2NO_2+2SO_2+96H_2O$

Mass of coal burned = $1.00\times 10^6$ metric ton = 1.00\times 10^6\times 10^3 kg=10^9 kg[/tex]

1 metric ton = $10^3kg$

Molar mass of coal :

$135\times 12g/mol+96\times 1g/mol+9\times 16 g/mol+1\times 14 g/mol+1\times 32 g/mol=1906 g/mol=1.906 kg/mol$

1 g = 0.001 kg

Moles of coal ,n :$\frac{10^9 kg}{1.906 kg/mol}=5.247\times 10^8 mol$

If 2 moles of coal on combustion gives 270 moles of carbon dioxide than n moles of coal will give;

$5.247\times 10^8 mol\times \frac{270}{2}=7.083\times 10^{10} mol$ of carbon dioxide.

Molar mass of carbon dioxide gas = 44 g/mol = 0.044 kg/mol

Mass of $7.083\times 10^{10} mol$ of carbon dioxide:

$7.083\times 10^{10} mol\times 0.044 kg/mol=3.11\times 10^{11} kg$