If a mole of sodium chloride weighs approximately 58 grams, and you have 29 grams of sodium chloride, how many moles of sodium c
1 answer:
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Answer:
12.5 g
Explanation:
<em>12.5 g of the compound would be formed.</em>
First, let us look at the balanced equation of reaction.
3 moles of Mg is required to react with 1 mole of N2 to produce 1 mole of product.
<em>Recall that: mole = mass/molar mass</em>
9.03 g of Mg = 9.03/24.3 = 0.3716 mole
3.48 g of N2 = 3.48/28 = 0.1243 mole
Mole ratio of Mg/N2 = 3:1
<em>Hence, there is no limiting reactant.</em>
3 moles of Mg is required for 1 mole of product.
0.3716 mole of Mg will therefore require:
0.3716 x 1/3 = 0.1239 moles of product.
Molar mass of product = 100.9 g/mol
Mass of 0.1239 mole = mole x molar mass
= 0.1239 x 100.9 = 12.5 g
Answer:
a) The volume of the system is 3.062 Liters.
b) Percentage error results from assuming ideal-gas behavior is 9.36%.
Explanation:
a)
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas =
V = Volume of gas = ?
n = number of moles of gas = 1 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 100°C = 100+273K=373 K
Putting values in above equation, we get:
V = 3.062 L
The volume of the system is 3.062 Liters.
b)
Volume of the container = V' = 2.8 L
System volume = V = 3.062 L
Percentage error of volume:
To calculate the percentage error, we use the equation:
Percentage error results from assuming ideal-gas behavior is 9.36%.