what volume of co2 is produced at stp when 270g of glucose are consumed in the following reaction? c6h12o6 + 6o2(g) -> 6co2 (g) + 6h2o(l)

202 L

Explanation:

Step 1: Write the balanced equation

C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

270 g × 1 mol/180.16 g = 1.50 mol

Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose

The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol

Step 4: Calculate the volume of 9.00 moles of CO₂ at STP

The volume of 1 mole of an ideal gas at STP is 22.4 L.

9.00 mol × 22.4 L/mol = 202 L

most likely he is cheating on you

answer; i believe that the correct answer is (a) potassium(k)