Questions

What volume of air containing 21\% of oxygen by volume is required to completely burn 10 gram of Sulphur which has a purity level of 98\%?

What volume of air containing 21\% of oxygen by volume is required to completely burn 10 gram of Sulphur which has a purity level of 98\%?

5 × 22.4 = 112 litres. This is 21 \% of volume of air.

What volume of air containing 21\% of oxygen by volume is required to completely burn one kg of carbon that contains 15\% of non combustible substance?

∴ 1.86 L of air is required to burn 1 g of carbon completely.

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What is the volume of air at NTP containing 21 oxygen?

3200L
What volume of air at N.T.P containing 21 \% oxygen by volume is required to completely burn 1000 g of sulphur containing 4 \% incombustible matter? If O2 is 672 L, then air =10021×672L=3200L.

What volume of air is needed for the combustion of?

So volume of oxygen required for combustion of 2 litre CH4= 2 x2=4 L. Volume of oxygen required for combustion of 2.75 litre H2= 2.75 x 0.5= 1.375 L. Volume of oxygen required for combustion of 0.25 litre CO = 0.25 x 0.5= 0.125. So total volume of oxygen required = 4+1.375+0.125= 5.5 Litres.

What is the volume of air at 1atm and 273k?

533.33 L
∴Volume of air required at STP=22.4×5×10021=533.33 L.

What is the volume of air at NTP?

NTP – Normal Temperature and Pressure At these conditions, the volume of 1 mol of a gas is 24.0548 liters.

What volume of air in m3 is needed for the combustion?

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Generally, 9 to 10 m3 of air is required for consuming 1 liter of fuel by a four-stroke engine.

How much air is required for the complete combustion of 1m3 of natural gas?

For natural gas-fired burners, the stoichiometric air required is 9.4-11 ft.3 / 1.0 ft. of natural gas or approximately an air-to-gas ratio of approximately 10:1. In this case, there is an excess oxygen level of 2\%. In the combustion zone, it is challenging to measure excess air.

What is the volume of air at 1atm?

How much air is needed for a complete combustion diesel?

∴ Oxygen required from air for the complete combustion of fuel will be (2.66C + 8H + S – O) which can be written as 2.66C + 8 (H – O/8) + S, the term in the bracket being known as the available hydrogen.

What is the volume of air at 1 atm and 273 K?

The volume of air required is 533.33 Liter.