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Hoochie [10]
3 years ago
5

If the percentage yield is 92.0% how many grams of CH3OH can be made by reaction of 5.6 x 103g with 1.0 x 103 g

Chemistry
1 answer:
Marizza181 [45]3 years ago
5 0
Answer is: 5888 grams of CH₃OH.
Chemical reaction: CO + 2H₂ → CH₃OH.
m(CO) = 5,6·10³ g = 5600 g.
m(H₂) = 1,0·10³ g = 1000 g.
n(CO) = 5600 g ÷ 28 g/mol
n(CO) = 200 mol, limited reactant.
n(H₂) = 1000 g ÷ 2 g/mol.
n(H₂) = 500 mol.
from reaction: n(CO) : n(CH₃OH) = 1 : 1.
n(CH₃OH) = n(CO) = 200 mol.
m(CH₃OH) = 200 mol · 32 g/mol · 0,92.
m(CH₃OH) = 5888 g.

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Answer:

pH = 3.95

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It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

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