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# Chem help: STP?

I have exams soon and got given an old paper and don't know how to do this question.

A portable gas heater runs from butane gas cylanders, that each contain 500g of butane gas.

2C4h10 + 13O2 => 8co2 + 10h2O

a) calculate the volume of carbon dioxide gas produced when one complete cylander is used at 1) STP, 2) 20 degrees celcius and 1 atm pressure

b) Calculate the mass of oxygen gas required to allow for the complete combustion of this cylander of butane.

c) Suggest what might happen if this amount of oxygen was not available.

Pleas help I dont want to fail my exam because I dont get how to do this

### 2 Answers

- Anonymous1 decade agoFavorite Answer
1)stp

2 mole of c4h10 produce 8 mol of co2

1mol of co2 at stp = 22.4 L(mole concept)

1 mole of c4 h10 =itds mol ecular mass= 4x 12 + 10 = 58 g

so, 2 mole= 2x 58 = 116 g

116 g of butane produces = 8 x 22.4 L of co2

EACH CYLINDER CONTAINS 500 G BUTANE )

500 g of butane will produce = 8 x 22.4 x 500 / 116 =772.41 L

2) at 20 c and 1 atm

PV= NRT

1 X V = 8 X 0.0821 X (20+273)

{ MOLE OF CO2 IN RXN ARE 8 CHANGE T IN KELVIN}

V=192.44 L

2 MOLE OF BUTANE ARE BURNT IN PRESENCE OF 13 MOLE OF O2

2 X 58 G OF BUTANE IS BURNT BY = 13 X 32 G OF O2

500 G WILL BE BURNT BY = 13 X 32 X 500 / 116

= 1793.10 G OF O2

IF THIS MUCH AMOUNT IS NOT AVAILABLE THEN THE BUTANE WILL REMAIN UNBURNT.........

- Anonymous4 years ago
use the molar quantity of gases at STP to discover moles of CO 391 L @ 22.4 Litres / mol = 17.40 5 moles CO via the equation 7 mole CO+15H2 --> 7H2O + a million mole C7H16 17.40 5 moles CO produces a million/seventh as many moles of C7H16 = 2.494 moles of C7H16 use the molar mass of C7H16, to discover grams: 2.494 moles of C7H16 @ a hundred.20 g/mol = 249.86 grams of C7H16 your answer rounded to 3 sig figs is 250. grams of C7H16 (your answer became rounded to 3 sig figs, because of the fact, "391 L" had purely 3 sig fis exhibiting) in case you have any problems with something, do no longer hesitate to digital mail me