2K (s) + 2 H₂O (1) 2 КОН (s) + H₂ (9) How many liters of H₂ are created from the reaction of 20.0 g potassium (K) with an excess of water? ает.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Solve like the example attached.  Apply the significant figure rule to your final answer.

Molar masses for all compounds in the equation:
K:39.10 g/mol
H₂O: 18.02 g/mol
KOH: 56.11 g/mol
H₂: 2.02 g/mol
2K (s) + 2 H₂O (1)
200g k
po
para
2 КОН (s) + H₂ (g)
How many liters of H₂ are created from the reaction of 20.0 g potassium (K) with an excess of water?
плает.
Transcribed Image Text:Molar masses for all compounds in the equation: K:39.10 g/mol H₂O: 18.02 g/mol KOH: 56.11 g/mol H₂: 2.02 g/mol 2K (s) + 2 H₂O (1) 200g k po para 2 КОН (s) + H₂ (g) How many liters of H₂ are created from the reaction of 20.0 g potassium (K) with an excess of water? плает.
Chemistry Vocabulary
WORD WALL
moles
-
mass-volume stoichiometry
+2 50₂ NU
(4)
0₂
(
0.70 L
1. Convert mass of the given chemical to moles of
the given chemical using the molar mass of the
7. Comert moles of the given chemical to moles of
the desired chemical using mole ratio.
3. Convert modes of the desired chemical to volume
of the desired chemical using the molar volume
of the desired cremlel.
2 CaSO4)
man
How much oxygen (0,) is required to
react completely with 3.5 g CaQ?
CaO motor mass: 56.06 g/mol -
Iml 60
$6.00 g (20
6
1 mol C
22.410.
3 mal Cadi mola
-
= 0.70
LO
Transcribed Image Text:Chemistry Vocabulary WORD WALL moles - mass-volume stoichiometry +2 50₂ NU (4) 0₂ ( 0.70 L 1. Convert mass of the given chemical to moles of the given chemical using the molar mass of the 7. Comert moles of the given chemical to moles of the desired chemical using mole ratio. 3. Convert modes of the desired chemical to volume of the desired chemical using the molar volume of the desired cremlel. 2 CaSO4) man How much oxygen (0,) is required to react completely with 3.5 g CaQ? CaO motor mass: 56.06 g/mol - Iml 60 $6.00 g (20 6 1 mol C 22.410. 3 mal Cadi mola - = 0.70 LO
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