Question

7-5.324 L of a gas is collected at 25.0"C and 846.0 mmHg. When the gas is changed to STP conditions, what is the new volume? 8-Glucose, C6H12O6, is metabolized in living systems. If 18.0 g of glucose are reacted with 7.50 L of 02 at 1.00 atm and 37 °C. how many g of water are produced? C6H1206(s)+6O2(g) -> 6CO2(g) +6H20()

Answer 1

5) The new volume is calculated:

V2 = V1 * P1 * T2 / P2 * T1 = 5.324 L * 846 mmHg * 273 K / 760 mmHg * 298 K = 5.429 L

6) The moles of reagents are calculated:

n Glu = m / MM = 18 g / 180 g / mol = 0.1 mol

n O2 = P * V / R * T = 1 * 7.5 L / 0.082 * 310 = 0.3 mol

The limit reagent is O2, the mass of water produced is calculated:

m H2O = 0.3 mol O2 * (6 mol H2O / 6 mol O2) * (18 g H2O / 1 mol) = 5.4 g

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