How much hydrogen will be released when water interacts with 8.5 g of sodium-potassium alloy containing 54% sodium?

We start by writing down the reaction equations for the metal alloy with water:

2Na + H₂O = 2NaOH + H₂ ↑;

2K + H₂O = 2KOH + H₂ ↑.

We analyze both reaction equations and see that for 2 moles of metal (Na and K) in both reactions there is 1 mole of H2, which occupies at n. at. volume of 22.4 liters.

Find the masses of metals in 8.5 grams of alloy.

Let’s apply the formula:

w = (mmet. / malloy) * 100%.

We are looking for each metal:

mNa = 8.5 g * 54%: 100% = 4.59 grams.

mK = 8.5 g * (100% – 54%): 100% = 3.91 g.

Mr (Na) = 23 g / mol, then the mass of 2 moles of sodium will be 46 grams.

We get the proportion:

46 grams of Na – 22.4 liters of H2;

4.59 grams Na – x liters H2.

x = (22.4 l * 4.59 g) / 46 g = 2.235 l. H2.

Likewise for potassium.

Mr (K) = 39 g / mol, then the mass of 2 moles of K is 78 g.

We get the proportion:

78 grams of K – 22.4 liters of H2;

3.91 g K – y l H2.

y = (22.4 l * 3.91 g) / 78 g = 1.123 l H2.

Let’s calculate the volume of hydrogen:

V (H2) = x + y = 2.235 L + 1.123 L ≈ 3.36 L H2.