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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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An unbalanced equation between sodium metal and chloride can be shown as:

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2 years ago
An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga
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Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

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<em> </em>

Computing the values of the problem:

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710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

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