What volume of hydrogen, measured at 25 C and 1.00 atm can be produced by dissolving 5.0 g of magnesium in hydrochloric acid? (Write and balance the equation for the reaction first.)
To find the volume of hydrogen gas produced, we first need to write and balance the equation for the reaction between magnesium and hydrochloric acid.
The reaction is as follows:
Mg + 2HCl -> MgCl2 + H2
Now, we can calculate the volume of hydrogen gas produced.
Step 1: Calculate the moles of magnesium used:
Given mass of magnesium = 5.0 g
Molar mass of magnesium (Mg) = 24.31 g/mol
Moles of Mg = given mass / molar mass
= 5.0 g / 24.31 g/mol
Step 2: Calculate the moles of hydrogen gas produced:
From the balanced equation, we observe that 1 mole of Mg reacts to produce 1 mole of H2 gas.
So, Moles of H2 = Moles of Mg
Step 3: Use the ideal gas law to calculate the volume of gas:
The ideal gas law equation is:
PV = nRT
Given conditions:
Pressure (P) = 1.00 atm
Temperature (T) = 25°C = 298 K (convert to Kelvin)
Let's assume the gas occupies the entire volume, so the volume (V) is what we need to find.
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K).
Rearranging the ideal gas law equation, we have:
V = (nRT) / P
Substituting the known values:
V = (moles of H2 * R * T) / P
= (moles of H2 * 0.0821 L·atm/(mol·K) * 298 K) / 1.00 atm
Now, plug in the value for the moles of H2 that we previously calculated to find the volume of hydrogen gas produced.