Answer:
The molar mass of the vapor is 43.83 g/mol
Explanation:
Given volume of gas = V = 247.3 mL = 0.2473 L
Temperature = T = 100
= 373 K
Pressure of the gas = P = 745 mmHg (1 atm = 760 mmHg)
Mass of vapor = 0.347 g
Assuming molar mass of gas to be M g/mol
The ideal gas equation is shown below
The molar mass of the vapor comes out to be 43.834 g/mol
Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol
The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
Answer:
Please refer below to the attached file
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To find the molarity of the compound, simply determine the molar mass of MgCl2 and then convert 50 g to moles using the molar mass of the compound. Then convert 150 ml to L = 0.15 L
Then divide the moles amount by the volume in L.
