Consider the following chemical equation.
MnCl2(s) + H2SO4(aq) → MnSO4(aq) + 2 HCl(g)
How many grams of manganese(II) chloride must react with sulfuric acid to release 50.0 mL of hydrogen chloride at STP?
Same process.
66.7 g
To determine the number of grams of manganese(II) chloride required to produce 50.0 mL of hydrogen chloride at STP, we need to follow these steps:
Step 1: Convert the volume of hydrogen chloride gas to moles.
Since the volume of a gas is given at STP (Standard Temperature and Pressure), we can use the ideal gas law to convert the volume to moles.
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
At STP, the pressure (P) is 1 atm, and the temperature (T) is 273 K.
Given:
Volume (V) = 50.0 mL = 0.0500 L
Pressure (P) = 1 atm
Temperature (T) = 273 K
Using the ideal gas law equation: PV = nRT, we can rearrange it to solve for n:
n = PV / RT
Plugging in the values, we get:
n = (1 atm) * (0.0500 L) / (0.0821 L.atm/mol.K * 273 K)
Step 2: Calculate the number of moles of hydrogen chloride.
By calculating the number of moles of hydrogen chloride, we can determine the amount of manganese(II) chloride required.
Step 3: Determine the molar ratio between hydrogen chloride and manganese(II) chloride.
By examining the balanced chemical equation: MnCl2(s) + H2SO4(aq) → MnSO4(aq) + 2 HCl(g), we see that for every molecule of MnCl2, we produce two molecules of HCl. This gives us the molar ratio between HCl and MnCl2.
Step 4: Convert moles of hydrogen chloride to moles of manganese(II) chloride.
Since we have the molar ratio between HCl and MnCl2, we can use it to convert moles of HCl to moles of MnCl2.
Step 5: Convert moles of manganese(II) chloride to grams.
Finally, using the molar mass of MnCl2, we can convert the moles of MnCl2 to grams.
Now, let's perform the calculations.
To solve this problem, we need to use stoichiometry and convert the given volume of hydrogen chloride gas to the mass of manganese(II) chloride.
1. Start by balancing the chemical equation:
MnCl2(s) + H2SO4(aq) → MnSO4(aq) + 2 HCl(g)
2. Calculate the molar volume of a gas at STP (standard temperature and pressure):
The molar volume of a gas at STP is 22.4 L/mol.
3. Convert the given volume of hydrogen chloride (HCl) to moles:
50.0 mL * (1 L / 1000 mL) * (1 mol / 22.4 L) = x moles of HCl
4. Use the stoichiometry of the balanced equation to find the moles of MnCl2:
From the balanced equation, we can see that for every 1 mole of HCl, 1 mole of MnCl2 reacts. Therefore, the moles of MnCl2 is the same as the moles of HCl (x moles).
5. Determine the molar mass of MnCl2:
Molar mass of MnCl2 = Molar mass of Mn + 2 * Molar mass of Cl
Molar mass of MnCl2 = 54.93805 g/mol + 2 * 35.453 g/mol
Molar mass of MnCl2 = 125.86805 g/mol
6. Calculate the mass of MnCl2 needed:
Mass of MnCl2 = x moles * Molar mass of MnCl2
Mass of MnCl2 = x moles * 125.86805 g/mol
Plug in the value of x (moles of HCl) that you calculated in step 3 to find the mass of MnCl2.