Question

CHEM 1111 Umt 6 I Knewton
X
+
& knewton.com/learn/course/05676594 2957-4c2b-affa Bf6e68dbb121/assignment/10045903 4d7b-4d1c-83495204525fa3e
1104
105
Ilios | 109
Ra
RE
||110
Db
1113
Sg Bh Hs
Ds
Cn
Nh
103
107
Fr
Mt
114
Ceบ
Rg
1115
Md
DARS
DER
SENS
117
TS
REA
Mes
Ly
Creo
041
Og
La
Ce
nis
Pr
Nd
Pm Sm
Eu
64
Gd
1922
Tb Dy
LA1
Но
77
Tm Yb
Er
10
111
19
La
>
Ас
2
Pa
Np
Pu
es
Am Cm
ta
Вk
Сf
2
Es
102
Md
200
Fm
B01
15
What volume of hydrogen at 300.0 K and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of
hydrochloric acid?
2 Ga(8) + 6 HCl(aq) → 2 GaCl, (aq) + 3 H,(9)
• Use 0.08206 Laim for the ideal gas constant.
• Recall that 1 atm = 760 torr.
• Round the answer to three significant figures.
mol K
Provide your answer below:
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10:59 PM
2021-02 15
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Answer 1

Answer:

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Answer 2

Answer: 4.94 L

Explanation:

To convert from the mass of gallium to the volume of H2(g), we need to do something like the following.

Mass of Ga ⟶ Moles of Ga ⟶ Moles of H2 (g) ⟶ Volumr og H2 (g)

The first two conversions are as shown below.

8.88g Ga× (1mol Ga / 69.723g Ga) × (3mol H2 / 2mol Ga) = 0.191mol H2

The pressure is given as 723 torr. To use the given value for the ideal gas constant (R=0.08206L atmmol K), we must convert this value to units of atm.

723torr × 1atm / 760torr = 0.9513atm

Finally, use the ideal gas law.

VH2= (nRT / P)H2 = (0.191mol × 0.08206L atm/mol K ×300.0 K) 0.9513atm = 4.943L

Therefore, after rounding to three significant figures, we find that the volume of hydrogen gas is about 4.94L.