Sorry I came a lil late,
The answer to your question is, 2.
Hope this helps! :)
Answer:
The correct answer is 0.12 grams.
Explanation:
The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.
Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.
The number of moles or n can be determined by using the equation, mass/molar mass.
R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).
On putting the values we get:
n = PV/RT
= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)
= 0.0043447 mol
Therefore, mass of CO will be moles * molar mass of CO
= 0.0043447 mol * 28 g/mol
= 0.12 g
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Increasing temperatures in a reaction increases the kinetic energy of the reactant molecules. This causes them to move fast and hence collide with a higher frequency. The higher the rate of collision between the molecules, the faster the reaction.
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