Calculate how many grams of lead (II) oxide can reduce the hydrogen obtained by the interaction of 460 g

| January 15, 2021 | education

Calculate how many grams of lead (II) oxide can reduce the hydrogen obtained by the interaction of 460 g of metallic sodium with water.

When sodium interacts with water, hydrogen is released and sodium hydroxide is formed. The resulting hydrogen reacts with lead (II) oxide to form water and pure lead metal.
The following reactions occur:
2 Na + 2 H2O = 2 NaOH + H2 [1]
PbO + H2 = Pb + H2O [2]
Molar masses of substances:
M (H2) = 2 g / mol
M (Na) = 23 g / mol
M (PbO) = 223 g / mol
The first step is to calculate the mass of hydrogen formed in the reaction [1]. Let’s make a proportion according to the reaction equation.
460 g of sodium corresponds to X g of hydrogen, as
2 * 23 g / mol sodium corresponds to 2 g / mol hydrogen
X = (460 * 2) / (2 * 23) = 20 g of hydrogen will be released.
Now let’s find the mass of lead (II) oxide. To do this, let’s make a proportion according to the reaction equation [2]:
Y g. PbO corresponds to 20 g. Hydrogen, as
223 g / mol PbO corresponds to 2 g / mol hydrogen
Y = (20 * 223) / 2 = 2230 g.
Answer: 2230 g of lead (II) oxide.



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