.How many moles of O2(g) are needed to react completely with 52.0 L of CH4(g) at STP to produce CO2(g) and H2O(g) according to the following reaction: CH4(g) + 2O2  CO2(g) + 2H2O(g)

4.64 moles of O₂ are needed.

Explanation:

Determine the balance reaction:

CH₄(g) +  2O₂(g)  →  CO₂(g)  +  2H₂O(g)

We verify the amount of methane at STP by the Ideal Gases Law

P . V= n . R . T

STP → 1 atm of pressure, 273K of T° (Standard conditions)

1 atm . 52L = n . 0.082 . 273K

(1 atm . 52L) / (0.082 . 273K) = n

52 / (0.082 . 273K) = n → 2.32 moles of CH₄

1 mol of methane reacts with 2 moles of oxygen

Then, 2.32 moles of methane must react with 4.64 moles of O₂ (2.32 .2) /1

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg

there are two significant figures.