In the following reaction: 2 K + 2 H2O → 2 KOH + H2, if we start with 9.0 grams of H2O, how many moles of H2 will be produced?
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Potassium reacts with water to produce potassium hydroxide (KOH) and hydrogen gas (H2) according to the equation:
2 K + 2 H2O --> 2 KOH + H2
Two moles of potassium react with 2 moles of water to form two moles of potassium hydroxide and 1 mole of hydrogen gas. The number of moles of hydrogen gas produced is half the number of moles of water and potassium that react together.
The molar mass of water is 18 g/mole. 9 g of water is half a mole of water. If the reaction is conducted with 9 g of water, half of mole of water reacts with half a mole of potassium to produce 1/4 mole of hydrogen.
The number of moles of hydrogen gas (H2) produced is 0.25 moles.
here the balanced reaction is
2 K + 2 H2O → 2 KOH + H2
We have 9.0 grams of water. We know that the molar mass of water is 18 g/mol.
Moles = mass/molar mass
Moles of water = 9.0 g/ (18g/mol) = 0.5 mol H2O
If we look at the reaction, 2 moles of water produces 1 mole of Hydrogen gas. And we have 0.5 moles of water. That is the molar ratio is 2 mol H2O : 1mol H2
0.5 mol H2O * ( 1mol H2/2 mol H2O)
0.25 moles of H2
Hence 9 g of water will produced 0.25 mol of H2 gas.
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