Question

The chemical formula of switch grass is determined as CH(1.488)O(0.666)N(0.016)S(0.001). If the ER value is 0.3 for gasifying the switch grass with air. What is the mass ratio of switch grass to air into a gasifier?

Answer 1

Theory:We need to stochimetrically balance the reaction and then accordingly calculate the mass ratio.

Solution Step 1:

Lets assume 1 mole of Swith grass

Molecular weight of swith grass CH(1.488)O(0.666)N(0.016)S(0.001)= 1.488*(12+1) + 0.666*16 +0.016*14 + 0.001*32 = 30.256 gm

Step 2: Balance reaction of Switch Grass:

CH(1.488)O(0.666)N(0.016)S(0.001) + XO2 --> 1.488CO2 + (1.488/2)H2O + 0.016NO2 + 0.001SO2

So baalnceinG O2, X = (1.488*2+1.488/2 + 0.016*2+0.001*2)/2 = 1.877 moles

So for 1 mole of switch grass, we need 1.877 moles of O2

But air has only 21% O2 so moles of air needed = 1.877/0.21 = 8.938 moles

Step 3: But actual fuel to air taio = 0.3

so for 1 mole of swith grass, moles of air = 8.938/0.3 = 29.79365 mole

molar mass of air = 29 gm/mol

so mass of air = moles*molar mass = 297.79365*29 = 864.016 gm

Step 4: Mass ratio = mass of switch grass/mass of air

= 30.256/864.016 = 0.035

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