Respuesta :
Answer:
The correct answer is option b.
Explanation:
[tex]2C_{135}H_{96}O_9NS+313O_2\rightarrow 270CO_2+2NO_2+2SO_2+96H_2O[/tex]
Mass of coal burned = [tex]1.00\times 10^6[/tex] metric ton = 1.00\times 10^6\times 10^3 kg=10^9 kg[/tex]
1 metric ton = [tex]10^3kg[/tex]
Molar mass of coal :
[tex]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[/tex]
1 g = 0.001 kg
Moles of coal ,n :[tex]\frac{10^9 kg}{1.906 kg/mol}=5.247\times 10^8 mol[/tex]
If 2 moles of coal on combustion gives 270 moles of carbon dioxide than n moles of coal will give;
[tex]5.247\times 10^8 mol\times \frac{270}{2}=7.083\times 10^{10} mol[/tex] of carbon dioxide.
Molar mass of carbon dioxide gas = 44 g/mol = 0.044 kg/mol
Mass of [tex]7.083\times 10^{10} mol[/tex] of carbon dioxide:
[tex]7.083\times 10^{10} mol\times 0.044 kg/mol=3.11\times 10^{11} kg[/tex]
