What volume of hydrogen gas is produced when 19.6 g of sodium reacts completely according to the following reaction at 25oC and 1 atm?
sodium (s) + water (l) sodium hydroxide (aq) + hydrogen (g)
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of hydrogen gas produced in this reaction, we need to use the concept of stoichiometry. Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction.
First, we need to balance the equation to determine the mole ratio between sodium and hydrogen gas in the reaction:
2 Na(s) + 2 H2O(l) -> 2 NaOH(aq) + H2(g)
From the balanced equation, we can see that 2 moles of sodium react to produce 1 mole of hydrogen gas.
Next, we need to convert the mass of sodium given (19.6 g) to moles. The molar mass of sodium (Na) is 22.99 g/mol, so we can calculate the number of moles of sodium:
moles of sodium = mass of sodium / molar mass of sodium
= 19.6 g / 22.99 g/mol
= 0.852 mol
Now, using the mole ratio from the balanced equation, we can determine the number of moles of hydrogen gas produced:
moles of hydrogen gas = moles of sodium * (1 mol H2 / 2 mol Na)
= 0.852 mol * (1 mol H2 / 2 mol Na)
= 0.426 mol
Lastly, we can use the ideal gas law to calculate the volume of the hydrogen gas at the given temperature and pressure. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in this case, 1 atm)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C = 298 K)
Rearranging the equation to solve for V:
V = (n * R * T) / P
= (0.426 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm
= 10.22 L
Therefore, the volume of hydrogen gas produced when 19.6 g of sodium reacts completely is approximately 10.22 liters.