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Butoxors [25]
3 years ago
9

What is the volume of 4.78g of O2 gas at STP?

Chemistry
2 answers:
My name is Ann [436]3 years ago
8 0

Answer:

Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that

1

mole of any ideal gas occupies

22.7 L

under STP conditions to calculate how many moles of gas you have in your sample.

Under STP conditions:

1 mole of an ideal gas = 22.7 L

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you know that your sample of gas occupies

2.28 L

under STP conditions, which are currently defined as a pressure of

100 kPa

and a temperature of

0

∘

C

.

This means that your sample will contain

2.28

L

⋅

molar volume of a gas at STP



1 mole gas

22.7

L

=

0.10044 moles gas

Now, the molar mass of the gas is the mass of exactly

1

mole of the gas. In your case, you know that you get

3.78 g

for every

0.10044

moles, which means that you have

1

mole

⋅

3.78 g

0.10044

moles

=

37.6 g

Since this is the mass of

1

mole of gas, you can say that the molar mass of the gas is

molar mass = 37.6 g mol

−

1

−−−−−−−−−−−−−−−−−−−−−−−

Nonamiya [84]3 years ago
6 0
I’m not surreeeeeeeeeee
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Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.1 g of butane is m
m_a_m_a [10]

Answer:

12.44 g

Explanation:

2C4H10 + 13O2 = 8CO2 + 10H2O

n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).

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Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.

mass of CO2 produced =

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M = 12.44 of CO2

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