a mol of Methane is 12 + 4 = 16 grams
so we used 256/16 = 16 mols
which is 16 * 6.022*10^23 molecules
we get 3 H2 molecules for every CH4 molecule
so we get 3 * 16 * 6.022 * 10^23 molecules of H2
1.Hydrogen gas can be made by reacting methane (CH4) with high temperature steam: CH4(g) + H2O(g) ----- CO(g) + 3H2(g)
How many hydrogen molecules are produced when 256 grams of methane reacts with steam?
2. Ammonia (NH3) reacts with oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O). Write and balance the chemical equation. How many liters of NO are produced when 2.0 liters of oxygen reacts with ammonia?
3.Phosphoric acid reacts with sodium hydroxide: H3PO4(aq) + 3NaOH(aq) --- Na3PO4(aq) + H2O(l)
If 3.50 mol H3PO4 is made to react with 10.0 mol NaOH, identify the limiting reagent
please help
3 answers
Now you try the second one.
1.Hydrogen gas can be made by reacting methane (CH4) with high temperature steam: CH4(g) + H2O(g) ----- CO(g) + 3H2(g)
How many hydrogen molecules are produced when 256 grams of methane reacts with steam?
mols CH4 = g/molar mass = ?
mols H2 = mols CH4 x (3 mols H2/1 mol CH4) = ?
Each mol H2 will have 6.022E23 molecules H2.
2. I assume you mean the gases are at STP. If so the problem is done the same way except the conversion factor for mols gas @ STP is 22.4 L = 1 mol
3. Done the same way as 1 but do it twice; i.e., one with mols H3PO4 to moles H2O produced. The second time do it with mols NaOH to mols H2O produced. In limiting reagent problems the correct value of mols H2O produced is the SMALLER number and the reagent responsible for that is the limiting reagent.
Post your work if you get stuck on any of these.
How many hydrogen molecules are produced when 256 grams of methane reacts with steam?
mols CH4 = g/molar mass = ?
mols H2 = mols CH4 x (3 mols H2/1 mol CH4) = ?
Each mol H2 will have 6.022E23 molecules H2.
2. I assume you mean the gases are at STP. If so the problem is done the same way except the conversion factor for mols gas @ STP is 22.4 L = 1 mol
3. Done the same way as 1 but do it twice; i.e., one with mols H3PO4 to moles H2O produced. The second time do it with mols NaOH to mols H2O produced. In limiting reagent problems the correct value of mols H2O produced is the SMALLER number and the reagent responsible for that is the limiting reagent.
Post your work if you get stuck on any of these.
1 answer