Ydrogen and fluorine combine according to the equation h2 (g) + f2 (g) → 2 hf (g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is

Molar mass (H2) = 2*1.0=2.0 g/mol

Molar mass (F2) = 2*19.0=38.0 g/mol

Molar mass (HF) = 1.0+19.0=20.0 g/mol

5.00 g H2 * 1mol H2 / 2 g H2=2.50 mol H2

38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2

H2 (g) + F2 (g) → 2 HF (g)

From reaction 1 mol 1 mol

From problem 2.50 mol 1.00mol

We can see that excess of H2, and that F2 is a limiting reactant.

So, the amount of HF is limited by the amount of F2.

H2 (g) + F2 (g) → 2 HF (g)

From reaction 1 mol 2 mol

From problem 1.00 mol 2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed