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3 years ago

How many grams of CaCo3 are required to prepare 50.0g of CaO

1 answer:

7 0

Answer is: 89 g of CaCO₃ is equired to prepare 50.0g of CaO.
Chemical reaction: CaCO₃ → CaO + CO₂.
m(CaO) = 50,0 g.
m(CaCO₃) = ?
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 50,0 g ÷ 56 g/mol.
n(CaO) = 0,89 mol.
from reaction: n(CaCO₃) : n(CaO) = 1 : 1.
n(CaCO₃) = n(CaO).
n(CaCO₃) = 0,89 mol.
m(CaCO₃) = m(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 0,89 mol · 100 g/mol
m(CaCO₃) = 89 g.
n - amount of substance.

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4 years ago

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

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Answer:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

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Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

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3 years ago

Calculate the number of moles of NaOH contained in 250. mL of a 0.05M solution.

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Answer:

0.0125mol

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Molarity (M) = number of moles (n) ÷ volume (V)

n = Molarity × Volume

According to this question, a 0.05M solution contains 250 mL of NaOH. The volume in litres is as follows:

1000mL = 1L

250mL = 250/1000

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n = 0.05 × 0.250

n = 0.0125

The number of moles of NaOH is 0.0125mol.

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3 years ago