Answer:
ok first your in my class and second im stuck too
Explanation:
Answer:
P = 2.145kPa
Explanation:
Mass = 22.1g
Molar mass of CO2 = 44g/mol
Vol = 165mL = 0.165L
T = -188°C = (-188 + 273.15)K = 85.15K
R = 8.314J/mol.K
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles if the gas
R = ideal gas constant
T = temperature of the gas
n = number of moles
n = mass / molar mass
n = 22.1 / 44 = 0.50moles
PV = nRT
P = nRT/ V
P = (0.5 × 8.314 × 85.15) / 0.165
P = 2145.26Pa = 2.145kPa
Pressure of the gas is 2.145kPa
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
<span>the theoretical yield which is the expected yield and the actual yield obtained are not always the same. therefore percent yield is calculated which shows how much of the percentage of the theoretical yield is actually obtained.
the theoretical yield = 56.0 g
actual yield = 47.0 g
percent yield = actual yield / theoretical yield x 100 %
percent yield = 47.0 / 56.0 x 100% = 83.9 %
percent yield = 83.9 %</span>