What volume of h2 gas (in l), measured at 756 mmhg and 90 ∘c, is required to synthesize 23.0 g ch3oh?

First let us calculate for the moles of CH3OH formed:

moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol

We see that there are 2 moles of H2 per mole of CH3OH, so:

moles H2 = 0.71875 mol * 2 = 1.4375 mol

Assuming ideal gas behaviour, we use the formula:

PV = nRT

V = nRT / P

V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15 K) / 756 mm Hg

V = 43.06 Liters

an obstruction

your answer is, orbit around the nucleus