What mass and how much of a substance of water is formed if 56 liters of hydrogen are used with O2?

To solve this problem, we write down given: V (H2) = 56 l
Find: the mass and amount of the substance of the resulting water.
Decision:
Let us write the equation for the reaction of the interaction of hydrogen and oxygen.
H2 + O2 = H2O
Let’s arrange the coefficients.
2H2 + O2 = 2H2O
Over hydrogen we write 56 l, and under hydrogen a constant molar volume equal to 22.4 l / mol. Since 2 mol of hydrogen entered, 2 mol * 22.4 l / mol = 44.8 l is needed.
Let’s calculate the molar mass of water:
M (H2O) = 1 * 2 + 16 = 18 g / mol
We write x g above the water, and 36 g under water.
Let’s compose and solve the proportion:
x = 56 * 36 / 44.8 = 45 g.
Let’s write down the formula for calculating the amount of substance:
v = m / M
v (H2O) = 45/18 = 2.5 mol.
Answer: m (H2O) = 45 g; v (H2O) = 2.5 mol.

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