Question

hua

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C 10. Given that Pb + HNO3 → Pb(NO3)2 + NO + H₂O (unbalanced). How many mls of NO
#
can be formed if 50.0 g of lead and 6.3 g of 20% nitric acid react under S.T.P?
(Pb=207)
(A) 11.2 (B) 22.4 (C) 112 (D) 224
a Pb + b HNO3 →→ cPb(NO3)₂ + ¿Not e H₂0
Pb: α = c = 1
a
H = b = 2e=2+d
N = b = 2 ctd = 2 td
0 = 3b = 6c+d²e = XAXte=1438
8
Pb + 1/{HNO3
二
X 3
⇒ 3Pb + 8HNO3
{ a = 1; c=[@] d = ² , e = {};b = { };
4
ze=2+d
ė = 2d²
50.0
2017.
à Pb + b HNO3 → cPb(NO3)₂ + ¿Not e H₂0
4
→ Pb(NO3)₂ + NO + H₂D
3
³Pb(NO3)₂ + 2 NO + 4 H ₂0
20% =
= 0.24/mol
=
zd
= 4d=2+d; d =
WIFW.IN
X
6.3.
⇒ x = 1.26
-
1.26
63
= 0.02 mol

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Answer 1

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我就是我

HNO3是限量試劑，HNO3和NO係數比8:2
0.02mole×(2/8)×22.4=0.112L=112mL