112.00g of nitrogen gas is reacted with an excess or abundant amount of oxygen gas. The single product, a gas, is determined to have a mass of 431.92 g. Additional experiments showed that the product consisted of 2.0 moles.
a) how many grams of oxygen reacted?
b) what percent composition by mass of both the O2 and N2 is the new compound?
c) how many moles of oxygen atoms reacted?
d) how many moles of nitrogen reacted?
e) what is the emipirical formula?
f) the molar mass of this compound is between 200-220 g. whats the name of this new compound ?
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i came here for help with the same question. i need answers
To solve this problem, we can use stoichiometry and the concept of mole ratios. Let's break down the problem step-by-step to find the answers:
a) To find the mass of oxygen gas that reacted, we need to determine the difference between the initial mass of the nitrogen gas and the final mass of the product.
Initial mass of nitrogen (N2) = 112.00 g
Final mass of compound = 431.92 g
Mass of oxygen gas reacted = Final mass - Initial mass of nitrogen
Mass of oxygen gas reacted = 431.92 g - 112.00 g
Mass of oxygen gas reacted = 319.92 g
Therefore, 319.92 g of oxygen gas reacted.
b) To find the percent composition by mass of both O2 and N2 in the compound, we need to determine the mass of each element in the compound and calculate the percentage.
Mass of oxygen in the compound = Mass of O2 reacted = 319.92 g
Mass of nitrogen in the compound = Mass of N2 reacted = 112.00 g
Percent composition of oxygen = (Mass of oxygen / Mass of compound) * 100%
Percent composition of oxygen = (319.92 g / 431.92 g) * 100%
Percent composition of oxygen = 74.05%
Percent composition of nitrogen = (Mass of nitrogen / Mass of compound) * 100%
Percent composition of nitrogen = (112.00 g / 431.92 g) * 100%
Percent composition of nitrogen = 25.95%
Therefore, the compound has 74.05% composition of oxygen and 25.95% composition of nitrogen by mass.
c) To find the number of moles of oxygen atoms reacted, we can use the molar mass of oxygen and the mass of oxygen gas reacted.
Molar mass of oxygen (O2) = 32.00 g/mol (16.00 g/mol per oxygen atom)
Number of moles of oxygen atoms = Mass of oxygen / Molar mass of oxygen
Number of moles of oxygen atoms = 319.92 g / 32.00 g/mol
Number of moles of oxygen atoms = 9.998 mol
Therefore, approximately 10 moles of oxygen atoms reacted.
d) Since we have the number of moles of oxygen atoms reacted, we can use the balanced chemical equation to determine the moles of nitrogen reacted. However, the chemical equation is not given in the problem, so we cannot directly determine the moles of nitrogen reacted.
e) The empirical formula of a compound represents the simplest whole-number ratio of elements. To find the empirical formula, we need the relative number of atoms of each element in the compound.
Since we know the percent composition by mass for oxygen (74.05%) and nitrogen (25.95%), we can assume a 100 g sample of the compound.
Mass of oxygen in 100 g of compound = 74.05 g
Mass of nitrogen in 100 g of compound = 25.95 g
Now, we can convert the masses to moles using the molar masses of oxygen and nitrogen:
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
Number of moles of oxygen = 74.05 g / 16.00 g/mol
Number of moles of oxygen = 4.628 mol
Number of moles of nitrogen = Mass of nitrogen / Molar mass of nitrogen
Number of moles of nitrogen = 25.95 g / 14.01 g/mol
Number of moles of nitrogen = 1.851 mol
Divide the number of moles of each element by the smallest number of moles (1.851 mol in this case) to obtain the simplest whole-number ratio:
Empirical formula = N2O4
Therefore, the empirical formula of the compound is N2O4.
f) The molar mass of the compound is given to be between 200 g/mol and 220 g/mol. By calculating the molar mass of the empirical formula N2O4, we can see if it falls within this range.
Molar mass of N2O4 = (2 × molar mass of N) + (4 × molar mass of O)
Molar mass of N2O4 = (2 × 14.01 g/mol) + (4 × 16.00 g/mol)
Molar mass of N2O4 = 28.02 g/mol + 64.00 g/mol
Molar mass of N2O4 = 92.02 g/mol
Since 92.02 g/mol is within the given range of 200-220 g/mol, the compound has a molar mass between 200-220 g/mol.
As for the name of this new compound, please note that for precise identification, the compound's chemical name may require more information, such as the exact structure and functional groups present.