What mass of manganese (II) chloride must react with sulfuric acid to release 49.5 mL of hydrogen chloride gas at STP? Mg(s) + H2SO4(aq) -->MgSO4(aq) +2HCl(g)
Magnesium chloride? Hmmm, where is that? Recheck your problem
What mass of manganese (II) chloride must react with sulfuric acid to release 49.5 mL of hydrogen chloride gas at STP? MnCl2(s) + H2SO4(aq)-->MnSO4(aq) + 2HCl(g)
To determine the mass of manganese (II) chloride required to produce 49.5 mL of hydrogen chloride gas at STP, we need to use stoichiometry and the given balanced chemical equation.
First, let's analyze the balanced equation to determine the stoichiometric ratio between manganese (II) chloride and hydrogen chloride.
According to the balanced equation:
1 mole of magnesium chloride reacts to produce 2 moles of hydrogen chloride.
Now, let's convert the volume of hydrogen chloride gas to moles using the ideal gas law at STP (Standard Temperature and Pressure).
STP conditions are defined as a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atmosphere (1 atm).
Using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (49.5 mL = 0.0495 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (273.15 K)
Plugging in the values:
(1 atm) * (0.0495 L) = n * (0.0821 L·atm/mol·K) * (273.15 K)
n = (1 atm * 0.0495 L) / (0.0821 L·atm/mol·K * 273.15 K)
n ≈ 0.00202 mol
Since 1 mole of magnesium chloride reacts to produce 2 moles of hydrogen chloride, we need twice the amount of moles of magnesium chloride.
0.00202 mol * 2 = 0.00404 mol
Finally, we need to convert moles of magnesium chloride to grams. We can do this by using the molar mass.
The molar mass of magnesium chloride (MgCl2) is:
(1 atom of magnesium) + (2 atoms of chlorine) = 24.31 g/mol + (2 * 35.45 g/mol) = 95.21 g/mol
Mass = moles * molar mass
= 0.00404 mol * 95.21 g/mol
Mass ≈ 0.385 g
Therefore, approximately 0.385 grams of manganese (II) chloride must react with sulfuric acid to release 49.5 mL of hydrogen chloride gas at STP.