Respuesta :
Answer:
0.4951 grams of Zn have been consumed.
Explanation:
Pressure at which gas is collected over water = p = 732 Torr
Vapor pressure of the water = p' = 26.74 Torr
Pressure of hydrogen gas = P
p = P + p'
P = p - p' = 732 Torr - 26.74 Torr = 705.26Torr
P = [tex]\frac{705.26}{760} atm=0.928 atm[/tex]
1 atm = 760 Torr
Volume of the hydrogen gas = V = 201 mL = 0.201 L
1 mL= 0.001 L
Moles of hydrogen gas = n
Temperature at which hydrogen gas collected = T = 27°C = 27+273 K = 300 K
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]=\frac{0.928 atm\times 0.201 mL}{0.0821 atm L/ mol K\times 300 K}[/tex]
n = 0.007573 mol
[tex]Zn(s) + H_2SO_4(aq)\rightarrow ZnSO_4(aq) + H_2(g)[/tex]
According to recation , 1 mole of hydrgen gas was obtained from 1 mole of zinc,then 0.007573 moles of hydrgen gas will be obtained from;
[tex]\frac{1}{1}\times 0.007573 mol=0.007573 mol[/tex] of zinc
Mass of 0.007573 moles of zinc:
0.007573 mol × 65.38 g/mol = 0.4951 g
0.4951 grams of Zn have been consumed.
The mass of Zinc that has been consumed is 0.49 g
We'll begin by calculating the number of mole of H₂.
Volume (V) = 201 mL = 201 / 1000 = 0.201 L
Temperature (T) = 27 °C = 27 + 273 = 300 K
Pressure (P) = 732 – 26.74 = 705.26 torr = 705.26 / 760 = 0.928 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =?
PV = nRT
Divide both side by RT
n = PV / RT
n = (0.928 × 0.201) / (0.0821 × 300)
n = 0.00757 mole
- Next, we shall determine the mole of Zn needed to produce 0.00757 mole of H₂
Zn + H₂SO₄ —>ZnSO₄ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore,
0.00757 mole of Zn will also react to produce 0.00757 mole of H₂
- Finally, we shall determine the mass of Zn that has been consumed.
Mole of Zn = 0.00757 mole
Molar mass of Zn = 65 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.00757 × 65
Mass of Zn = 0.49 g
Thus, 0.49 g of Zn has been consumed
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