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artcher [175]
3 years ago
10

How many grams of butane (C H20) must be burned in an excess of O, to produce 15.0

Chemistry
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

Mass of butane = 1.87 g

Explanation:

Form a balanced chemical equation for the reaction stated:

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

Number of moles of CO₂

=  \frac{mass}{molar \: mass}

=  \frac{15}{12 + 2(16)}

= 0.34 \: mol

From the equation,

1 mol C₄H₁₀ : 4 mol CO₂

0.085 mol C₄H₁₀ : 0.34 mol CO₂

Mass of butane

= number \: of \: moles \times molar \: mass

= 0.085 \times (12 + 10(1))

= 1.87 \: g

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Answer:0.59M

Explanation:

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3 years ago
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Answer:

1. D   2. C   3. B

Explanation:

1. The product is always what is being yielded from the equation, or what is created. Since both 4H2O and 3CO2 are created from the reactants, they are both products of the reaction.

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2 years ago
CH3OH can be synthesized by the following reaction.
puteri [66]

Answer:

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

Explanation:

Step 1: Data given

CO(g)+2H2(g)?CH3OH(g)

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.04 g/mol

Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?

Pressure = 754 mmHg = 0.992 atm

Temperature = 90°C = 363 Kelvin

mass of CH3OH produced = 23.0 grams

Step 3: Calculate moles of CH3OH

Moles CH3OH = mass CH3OH / Molar mass CH3OH

Moles CH3OH = 23.0 grams / 32.04 g/mol

Moles CH3OH = 0.718 moles

Step 4: Calculate moles of H2

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO

Step 5: Calculate volume of H2

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 1.436 moles H2

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (1.436*0.08206*363)/0.992

V = 43.12 L

Step 6: Calculate volume of CO

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 0.718  moles CO

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (0.718*0.08206*363)/0.992

V = 21.56 L

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

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