What mass of manganese(II) chloride must react with sulfuric acid to release 54.0mL of hydrogen chloride gas at STP?
MnCl2(s)+H2SO4(aq)�¨MnSO4(aq)+2HCl(g)
From the equation, find the stoichiometric ratio of HCl and MnCl2 which equals 2.
This means that 1 mol of MnCl2 will generate 2 mol of HCl.
Using the fact that 1 mol of HCl occupies 22.4 L at STP, you will find how many mol of MnCl2 will be required to make 0.054 L of HCl.
To find the mass of manganese(II) chloride required, we need to use stoichiometry. The stoichiometry of the balanced equation tells us that 1 mole of MnCl2 reacts with 2 moles of HCl.
Step 1: Convert the given volume of HCl gas to moles.
To do this, we need to use the ideal gas law equation. At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 liters. Since we have 54.0 mL of HCl gas, we can convert it to liters by dividing by 1000: 54.0 mL ÷ 1000 = 0.054 L.
Using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (.0821 L·atm/mol·K), and T is the temperature in Kelvin.
At STP, the pressure (P) is 1 atm, and the temperature (T) is 273 K. Plugging in the values, we have:
(1 atm)(0.054 L) = n(.0821 L·atm/mol·K)(273 K)
Solving for n:
n = (1 atm)(0.054 L) ÷ (.0821 L·atm/mol·K)(273 K)
n ≈ 0.00198 moles
Step 2: Use stoichiometry to find the moles of MnCl2.
According to the balanced equation, 1 mole of MnCl2 reacts with 2 moles of HCl. Therefore, the number of moles of MnCl2 is half of the moles of HCl.
Number of moles of MnCl2 = 0.00198 moles HCl ÷ 2 ≈ 0.00099 moles
Step 3: Calculate the mass of MnCl2 using its molar mass.
The molar mass of MnCl2 is the sum of the atomic masses of manganese (55.0 g/mol) and two chlorine atoms (35.5 g/mol each).
Molar mass of MnCl2 = 55.0 g/mol + (2 * 35.5 g/mol) = 125.0 g/mol
Mass of MnCl2 = Number of moles of MnCl2 * Molar mass of MnCl2
Mass of MnCl2 = 0.00099 moles * 125.0 g/mol ≈ 0.124 g
Therefore, approximately 0.124 grams of manganese(II) chloride must react with sulfuric acid to release 54.0 mL of hydrogen chloride gas at STP.
Note: Make sure to round the final answer to the appropriate number of significant figures based on the given data.