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4 years ago

Write a balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product.

Chemistry

1 answer:

pentagon [3]4 years ago

7 0

Answer:

2 CH₃OH + 3 O₂ ⇒ 2 CO₂ + 4 H₂O

Explanation:

Methanol is CH₃OH. Oxygen is O₂. A combustion produces CO₂ and H₂O. Create an equation using this information and balance.

CH₃OH + O₂ ⇒ CO₂ + H₂O

2 CH₃OH + 3 O₂ ⇒ 2 CO₂ + 4 H₂O

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Alecsey [184]

The answer to your question is Coefficients.

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3 years ago

Read 2 more answers

What are the mole fraction and the mass percent of a solution made by dissolving 0.21 g KBr in 0.355 L water? (d = 1.00 g/mL.) m

IRISSAK [1]

Answer:

Mol fraction H2O = 0.99991

Mol fraction KBr = 0.00009

mass % KBr = 0.059 %

mass % H2O = 99.941 %

Explanation:

Step 1: Data given

Mass of KBr = 0.21 grams

Molar mass KBr = 119 g/mol

Volume of water = 355 mL

Density of water = 1.00 g/mL

Molar mass water = 18.02 g/mol

Step 2: Calculate mass water

Mass water = 355 mL * 1g /mL

Mass water = 355 grams

Step 3: Calculate moles water

Moles water = mass water / molar mass water

Moles water = 355 grams / 18.02 g/mol

Moles water = 19.7 moles

Step 4: Calculate moles KBr

Moles KBr = 0.21 grams / 119 g/mol

Moles KBr = 0.00176 moles

Step 5: Calculate total moles

Total moles = 19.7 moles + 0.00176 moles

Total moles = 19.70176 moles

Step 6: Calculate mol fraction

Mol fraction H2O = 19.7 moles / 19.70176 moles

Mol fraction H2O = 0.99991

Step 7: Calculate mol fraction KBr

Mol fraction KBr = 0.00176 / 19.70176

Mol fraction KBr = 0.00009

Step 6: Calculate mass %

mass % KBR = (0.21 grams / (0.21 + 355) grams) *100%

mass % KBr = 0.059 %

mass % H2O = (355 grams / 355.21 grams) *100%

mass % H2O = 99.941 %

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3 years ago

Identify the element using the Bohr model below.

Maurinko [17]

Nitrogen I believe . I need 20 characters.

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3 years ago

5. If 1 g of a gas occupies a volume of 300 mL at STP, what is the molecular weight of the gas?

ikadub [295]

Answer:

74,67 gr/mol

Explanation:

At STP 1 mole of an ideal gas has volume of 22,4 L. Since we know the volume of the gas we can find the number of moles of the gas. (300 mL=0,3 L)

n=0,3L/22,4 L=0,01339 mol

Since we know weight of the gas as 1 g, we can find the molecular weight as;

MW=1 g/0,01339 mol =74,67 gr/mol

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3 years ago

Which expression can be used to calculate the density of a 129.5-gram sample of bronze that has a volume of 14.8 cubic centimete

PilotLPTM [1.2K]

Answer:

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Explanation:

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2 years ago