# CH3COOH + Mg calculate the volume of hydrogen that will be obtained by reacting 48 g of Mg

**CH3COOH + Mg calculate the volume of hydrogen that will be obtained by reacting 48 g of Mg with acetic acid if the product yield (H) = 80% or 0.8.**

Let’s write down the reaction equation and arrange the coefficients:

Mg + 2CH3COOH = (CH3COO) 2Mg + H2

Let’s calculate the amount of magnesium substance that has reacted: n (Mg) = m (Mg) / M (Mg), where

m (Mg) is the mass of magnesium, M (Mg) is the molar mass of magnesium.

n (Mg) = 48/24 = 2 mol

It can be seen from the reaction equation that the amount of hydrogen substance is nt (H2) = n (Mg). This is a theoretical way out. Taking into account the practical output w: np (H2) = w * nt (H2) = 0.8 * 2 = 1.6 mol.

Let us find the volume at n.u .: V (H2) = nп (H2) * Vm, where Vm is the molar volume.

V (H2) = 1.6 * 22.4 = 35.84 L