What mass (g) of magnesium chloride (MgCl2) must react with sulfuric acid (H2SO4) to release 890.2 ml of Hydrogen chloride (HCl) gas at STP according to the following equation? Round to one decimal place.

MgCl2 (s) + H2SO4 --> MgSO4 (aq) + 2 HCl

To determine the mass of magnesium chloride (MgCl2) needed to release 890.2 ml of hydrogen chloride (HCl) gas at STP, we can follow these steps:

Step 1: Determine the volume of gas at STP.
STP stands for Standard Temperature and Pressure, which is defined as 0 degrees Celsius and 1 atmosphere of pressure. At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, we need to convert the given volume from milliliters to liters:
890.2 ml = 890.2/1000 = 0.8902 L

Step 2: Apply stoichiometry to find the number of moles of HCl produced.
From the balanced equation:
1 mole of MgCl2 produces 2 moles of HCl
Therefore, we can set up the following proportion:
1 mole MgCl2 / 2 moles HCl = x moles MgCl2 / 0.8902 moles HCl

Simplifying the proportion:
x = (1 mole MgCl2 / 2 moles HCl) * 0.8902 moles HCl
x = (0.4451 moles MgCl2)

Step 3: Convert moles of MgCl2 to grams.
The molar mass of MgCl2 can be calculated as follows:
Molar mass (MgCl2) = atomic mass (Mg) + (2 * atomic mass (Cl))
Molar mass (Mg) = 24.31 g/mol
Molar mass (Cl) = 35.45 g/mol
Molar mass (MgCl2) = 24.31 + (2 * 35.45) = 95.21 g/mol

Now, we can determine the mass of MgCl2:
Mass (MgCl2) = moles (MgCl2) * molar mass (MgCl2)
Mass (MgCl2) = 0.4451 moles * 95.21 g/mol = 42.09 g

Therefore, the mass of magnesium chloride (MgCl2) needed to produce 890.2 ml of hydrogen chloride (HCl) gas at STP is approximately 42.1 grams (rounded to one decimal place).

To find the mass of magnesium chloride (MgCl2) required to produce a certain volume of hydrogen chloride (HCl) gas at STP, we need to follow these steps:

Step 1: Convert the given volume of HCl gas to the number of moles.
- We are given the volume of HCl gas as 890.2 mL.
- To convert to moles, we need to use the ideal gas law equation: PV = nRT.
- Since the gas is at standard temperature and pressure (STP), we can use the values:
- P = 1 atm
- T = 273 K (0 °C)
- R = 0.0821 L·atm/mol·K (the ideal gas constant)
- Rearranging the ideal gas law equation, we have: n = PV / RT.
- Plugging in the values, we get: n = (1 atm) * (890.2 mL / 1000 mL/L) / (0.0821 L·atm/mol·K * 273 K).
- Simplifying, we find: n ≈ 0.0360 moles.

Step 2: Determine the molar ratio between MgCl2 and HCl.
- From the balanced equation, we can see that for every 1 mole of MgCl2, 2 moles of HCl are produced.
- This means that the molar ratio of MgCl2 to HCl is 1:2.

Step 3: Calculate the molar mass of MgCl2.
- The molar mass of magnesium (Mg) is approximately 24.3 g/mol.
- The molar mass of chloride (Cl) is approximately 35.5 g/mol.
- Since MgCl2 consists of one magnesium atom and two chloride atoms, we can calculate the molar mass:
- MgCl2 = (1 * 24.3 g/mol) + (2 * 35.5 g/mol).
- Simplifying, we find: MgCl2 = 95.3 g/mol.

Step 4: Convert moles of MgCl2 to grams.
- We know that 1 mole of MgCl2 weighs approximately 95.3 grams.
- From Step 2, we found that the molar ratio of MgCl2 to HCl is 1:2.
- Therefore, if 1 mole of MgCl2 produces 2 moles of HCl, then 0.0360 moles of MgCl2 will produce (2 * 0.0360) moles of HCl.
- Converting moles of MgCl2 to grams, we have: mass = moles * molar mass.
- Plugging in the values, we get: mass ≈ 0.0360 * 95.3 g ≈ 3.43 g.

Therefore, approximately 3.4 grams of magnesium chloride (MgCl2) must react with sulfuric acid (H2SO4) to release 890.2 mL of hydrogen chloride (HCl) gas at STP.