Answer:
There is 0.0677 grams of H2 gas obtained
Explanation:
Step 1: Data given
The total pressure (988 mmHg) is the sum of the pressure of the collected hydrogen + the vapor pressure of water (17.54 mmHg).
ptotal = p(H2)+ p(H2O)
p(H2) = ptotal - pH2O = 988 mmHg - 17.54 mmHg = 970.46 mmHg
Step 2: Calculate moles of H2 gas
Use the ideal gas law to calculate the moles of H2 gas
PV = nRT
n = PV / RT
⇒ with p = pressure of H2 in atm = 970.46 mmHg * (1 atm /760 mmHg) = 1.277 atm
⇒ V = volume of H2 in L = 641 mL x (1 L / 1000 mL) = 0.641 L
⇒ n = the number of moles of H2 = TO BE DETERMINED
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 20.0 °C = 293.15 Kelvin
n = (1.277)(0.641) / (0.08206)(298.15) = 0.0335 moles H2
Step 3: Calculate mass of H2
Mass of H2 = moles H2 ¨molar H2
0.0335 moles H2 * 2.02 g/mol H2 = 0.0677g H2
There is 0.0677 grams of H2 gas obtained
The mass of hydrogen gas obtained is 0.068 g of hydrogen gas.
The equation of the reaction is;
Ca(s) +2H2O(l) →Ca(OH)2(aq) + H2(g)
We have to obtain the number of moles of hydrogen gas produced using the information in the question.
P = 988mmHg - 17.54 mmHg = 1.28 atm
V = 641 mL or 0.641 L
T = 20 + 273 = 293 K
n = ?
R = 0.082 atmLK-1mol-1
From;
PV = nRT
n = PV/RT
n = 1.28 atm × 0.641 L/ 0.082 atmLK-1mol-1 × 293 K
n = 0.82/24.026
n = 0.034 moles
Mass of hydrogen = 0.034 moles × 2 g/mol = 0.068 g of hydrogen gas
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