So would you say that less than 5.3 g of copper II is formed and some aluminum is left in the reaction mixture? Aluminium is the limiting reactant meaning all of it is consumed in the reaction. Only copper chloride is in excess and hence some of it remain in solution.
Answer:
Mass = 14.72 g
Explanation:
Given data:
Volume of hydrogen = 6.58 L
Temperature of gas = 32°C (32+273 = 305 k)
Pressure of gas = 895 mmHg (895/760 = 1.2 atm)
Mass of sodium required = ?
Solution:
Chemical equation:
2HCl + 2Na → 2NaCl + H₂
Number of moles of hydrogen:
PV = nRT
1.2 atm × 6.58 L = n× 0.0821 atm.L/mol.K×305 K
7.9 atm.L = n× 25.0 atm.L/mol
n = 7.9 atm.L / 25.0 atm.L/mol
n = 0.32 mol
Now we will compare the moles of hydrogen with sodium.
H₂ : Na
1 : 2
0.32 : 2/1×0.32 = 0.64 mol
Mass of sodium:
Mass = number of moles × molar mass
Mass = 0.64 mol × 23 g/mol
Mass = 14.72 g
The empirical formula for this would be FeS
C - straight line
.....................................................................................
