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
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.
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
Now you try the second one.
1. To find out how many hydrogen molecules are produced when 256 grams of methane reacts with steam, we need to first determine the number of moles of methane.
The molar mass of methane (CH4) is:
(1 x atomic mass of Carbon) + (4 x atomic mass of Hydrogen) = 12.01 + (4 x 1.01) = 16.05 g/mol
Now we can calculate the number of moles of methane:
Number of moles = Mass / Molar Mass
Number of moles = 256 g / 16.05 g/mol ≈ 15.97 mol
From the balanced equation, we can see that for every 1 mole of methane, 3 moles of hydrogen molecules are produced. Therefore, the number of moles of hydrogen molecules produced can be calculated as:
Number of moles of hydrogen = Number of moles of methane x (3 moles of H2 / 1 mole of CH4)
Number of moles of hydrogen = 15.97 mol x (3 moles of H2 / 1 mole of CH4)
Finally, we can convert the number of moles of hydrogen to molecules by multiplying by Avogadro's number (6.022 x 10^23 molecules/mol):
Number of hydrogen molecules = Number of moles of hydrogen x Avogadro's number
Number of hydrogen molecules = 15.97 mol x ( 3 moles of H2 / 1 mole of CH4) x (6.022 x 10^23 molecules/mol)
2. To write and balance the chemical equation for the reaction between ammonia and oxygen, we need to consider the following:
NH3(g) + O2(g) → NO(g) + H2O(g)
Now, we need to balance the equation by adjusting the coefficients:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
To determine the number of liters of NO produced when 2.0 liters of oxygen reacts with ammonia, we need to convert the volume of oxygen to moles using the ideal gas law equation:
PV = nRT
We can rearrange the equation to solve for the number of moles:
n = PV / RT
Assuming standard temperature and pressure (STP), we have:
P = 1 atm
V = 2.0 liters
R = 0.0821 L•atm/(K•mol)
T = 273 K
n = (2.0 L x 1 atm) / (0.0821 L•atm/(K•mol) x 273 K)
Once we have the number of moles of oxygen, we can use the stoichiometric coefficients from the balanced equation to find the number of moles of nitrogen monoxide produced.
Finally, we can convert the number of moles of nitrogen monoxide to liters using the ideal gas law equation.
3. To identify the limiting reagent in the reaction between phosphoric acid and sodium hydroxide, we need to compare the stoichiometric ratios of the reactants.
From the balanced equation: H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + H2O(l)
We can see that the stoichiometric ratio between H3PO4 and NaOH is 1:3. This means that 1 mole of H3PO4 reacts with 3 moles of NaOH.
Given that we have 3.50 mol of H3PO4 and 10.0 mol of NaOH, to determine the limiting reagent, we calculate the moles of NaOH that would be needed to react completely with the available H3PO4:
Moles of NaOH needed = (3.50 mol H3PO4) x (3 mol NaOH / 1 mol H3PO4)
If the moles of NaOH needed exceeds the actual moles of NaOH available (10.0 mol), then NaOH is the limiting reagent. If the moles of NaOH needed are less than or equal to the actual moles of NaOH available, then H3PO4 is the limiting reagent.