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How many moles of co2 form when 58 grams of butane burn in oxygen?

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## Related Questions

###### Asked in Elements and Compounds

### How many grams of oxygen are required to burn 4.8 mol of butane?

The balanced equation for the reaction is 2 C4H10 + 13 O2 ->
8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are
required to burn 2 moles of butane. By proportionality, (4.8/2)13
or 31.2 moles of oxygen are required to burn 4.8 moles of butane.
This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.

###### Asked in Chemical Equations

### How many grams of o2 are needed to completely burn 96.6 grams of c3h8?

The chemical equation for this reaction is:
C3H8(g) + 5O2(g) (arrow) 4H2O(l) + 3CO2(g)
Using stoichiometry, 5 moles of oxygen are needed to burn 1 mole
of C3H8. 96.6 grams is equal to 0.5889 moles. You multiply this by
five to find the amount of moles of O2 required, which is 2.945
moles. Converting this to grams gives 94.24 grams of O2.

###### Asked in Metal and Alloys

### How many Grams of magnesium oxide are produced when 3 moles of magnesium burn?

A balanced equation for the reaction is 3 Mg + 3/2 O2 = 3 MgO.
the atomic mass of mass of magnesium is about 24.3 and that of
oxygen is about 16.0; therefore the mass of magnesium oxide
produced is 3 (24.3 + 16.0) = 120.9 grams. (If the specification of
"3 moles" of magnesium is considered to have only one significant
digit, this answer should be written instead as 1 X 102 grams.)

###### Asked in Magnesium

### What is the mass of magnesium oxide when you burn 7 grams of magnesium?

Mg grams -> (use Mg's molar mass) -> Mg moles -> (use
ratio of moles - use balanced equation) -> MgO moles -> (use
MgO's molar mass) -> grams MgO
set up the equation:
Mg + O2 --> MgO
(we know the product is MgO and not MgO2 because magnesium has a
charge of 2+ while oxygen has a charge or 2-)
balance the equation:
2Mg + O2 --> 2MgO
Molar mass of Mg: 24.31 g/mol
Molar mass of MgO: 44.30 g/mol (add the molar mass of Magnesium
- 24.31g/mol and the molar mass of Oxygen - 15.99g/mol
together)
(use periodic table to find these)
7.0 grams of Mg
To find the moles of Magnesium you use the molar mass of Mg.
(7.0 g Mg)*(1 mol Mg / 24.31 g Mg) =0.2879 moles Mg
notice how the grams cancel to leave you with moles - remember
dividing by a fraction is the same as multiplying by the
reciprocal
Now use the balanced equation's coefficients and the moles of Mg
to determine the number of moles of MgO present.
2Mg + O2 --> 2MgO
2 moles Mg : 2 moles MgO -> divide both sides by 2 and it
obviously becomes a 'one to one' ratio. This means that the number
of moles of Mg is equal to the number of moles of MgO. This means
that there are 0.2879 moles of MgO.
Now that we know MgO's molar mass and the number of moles of MgO
we have, the grams of MgO produced can be determined.
(0.2879 moles MgO)*(44.30 g MgO / 1 mol MgO) = 12.75 grams
MgO

###### Asked in Chemistry

### Does pure oxygen burn?

###### Asked in Elements and Compounds

### Can oxygen burn?

###### Asked in Elements and Compounds

### What weight of oxygen is required to completely burn 19.8 g of octane?

The chemical equation for complete burning of octane is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O.
This equation shows that 25 moles of diatomic oxygen are
required to completely burn each two moles of octane. The gram
molecular mass of octane is 114.23 and the gram molecular mass of
diatomic oxygen is 2(15.9994). Therefore, the ratio of the mass of
oxygen required to completely burn any given mass of octane to the
mass of the octane to be burned is 50(15.9994)/2(114.23) or 3.5016,
to the justified number of significant digits (the same as the
number of digits in the least precisely specified datum 114.23, and
the mass of oxygen required to burn 19.8 g of octane is
(3.5016)(19.8) or 69.3 grams to the justified number of significant
digits, now limited by the less precisely specified datum
119.8.