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kvv77 [185]
3 years ago
13

If 24.6 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 301 Kelvin and 1.01 atmosph

eres? Show all of the work used to solve this problem. 2Li (s) + 2H2O (l) yields 2LiOH (aq) + H2 (g)
Chemistry
1 answer:
andreev551 [17]3 years ago
7 0
Answer: 43.3 l


Explanation:

1) Chemical equation:
2 Li(s) + 2 H₂O (l) → 2LiOH(aq) + H₂ (g)

2) Mole ratios:

2 mol Li : 2 mol H₂O : 2 mol LiOH : 1 mol H₂

3) Number of moles of Li that react

n = mass in grams / atomic mass = 24.6g / 6.941 g/mol = 3.54 moles

4) Yield

Proportion:

2 mol Li / 1 mol H₂ = 3.54 mol  Li/ x


⇒ x = 3.54 mol Li × 1 mol H / 2 mol Li = 1.77 mol H₂

4) Ideal gas equation

PV = nRT ⇒ V = nRT / P

V = 1.77 mol × 0.0821 [atm×l / (mol×K)] × 301 K / 1.01 atm = 43.3 l


V = 43.3 l ← answer



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Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

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n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

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And the resulting pH is:

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