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Rzqust [24]
3 years ago
15

How many grams of o2 are required to produce 7.00 g of co2?

Chemistry
1 answer:
Anna71 [15]3 years ago
7 0
Hey there!:

Molar mass:

O2 = 31.99 g/mol
CO2 = 44 g/mol

By the stoichiometry of the reaction:

C + O2 = CO2

31.99 g O2 --------------- 44 g CO2
mass O2 ------------------ 7.00 g CO2

mass O2 = ( 7.00 * 31.99 ) / 44

mass O2 = 223.93 / 44

mass O2 = 5.089 g
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Seems to be an ideal gas law question. The relevant equation is:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.082 atm-L/mole-deg K), and T is temperature in Kelvins. STP means standard temperature and pressure and this is taken as 1 atm and 0º C or 273 K.

To calculate the number of molecules we will use the constant 6.023 * 10^23 molecules/mole and, therefore, we will need to know the number of moles (n). So, first we’ll rearrange the gas law equation, isolating ’n’ and then put the numbers in.

n = PV/RT = 1 * 1 / (0.082)(273) = 0.0447 moles

So, to calculate the number of molecules, multiple this by the number of molecules in a mole and you get:

# molecules of nitrogen in 1 Liter at STP = 6.023 * 10^23 molecules/mole * 0.0447 moles = 2.6905 * 10^22 molecules

Note, it does not matter what the gas is.

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