Calculate the number of grams of oxygen required to convert 48.0 g of glucose to co2 and h2o.

1 answer:

4 0

I believe that the balanced chemical reaction is:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

So the number of grams of oxygen required is:

mass O2 required = 48 g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32 grams O2 / 1 mole)

mass O2 required = 51.15 grams

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Answer:

2000 L

General Formulas and Concepts:

Atomic Structure

Gas Laws

Ideal Gas Law: PV = nRT

Explanation:

Step 1: Define

[Given] 8.8 moles gas

[Given] 0.12 atm

[Given] 56 °C = 329.15 K

Step 2: Solve for V

Substitute in variables [Ideal Gas Law Formula]: $\displaystyle (0.12 \ atm)V = (8.8 \ mol)(0.0821 \ \frac{L \cdot atm}{mol \cdot K})(329.15 \ K)$
Isolate V: $\displaystyle V = \frac{(8.8 \ mol)(0.0821 \ \frac{L \cdot atm}{mol \cdot K})(329.15 \ K)}{(0.12 \ atm)}$
Multiply/Divide [Cancel out units]: $\displaystyle V = 1981.7 \ L$