How much oxygen (in grams) will form from 44.2 mg of N2O5?
2N2O5(s) → 4NO2(g) + O2(g)
____× 10^-3(g) O2
N = 14 g/mol
O = 16 g/mol
2 Mols N2O5 = 56+160 = 216 grams
1 mol O2 = 32 grams
so 32 grams O2 for each 216 grams N2O5
so
(32/216)44.2
To calculate the amount of oxygen formed from 44.2 mg of N2O5, we need to use the ratio of oxygen to N2O5 from the balanced equation.
The molar mass of N2O5 is:
2 * 14.01 g/mol (for nitrogen) + 5 * 16.00 g/mol (for oxygen) = 108.01 g/mol
To find the moles of N2O5, we divide the given mass by the molar mass:
44.2 mg / 108.01 g/mol = 0.4091 mmol
From the balanced equation, we can see that for every 2 moles of N2O5, 1 mole of O2 is produced.
So, the moles of O2 formed can be calculated as:
0.4091 mmol N2O5 × (1 mol O2 / 2 mol N2O5) = 0.2046 mmol O2
Finally, convert the moles of oxygen to grams using the molar mass of oxygen (32.00 g/mol):
0.2046 mmol O2 × 32.00 g/mol = 6.55 mg O2
Therefore, 44.2 mg of N2O5 will produce 6.55 mg of O2.
To calculate the amount of oxygen (in grams) formed from a given mass of N2O5, you need to use stoichiometry.
First, calculate the molar mass of N2O5, which can be obtained by adding up the atomic masses of its constituent elements:
N2O5: 2 x (14.01 g/mol) + 5 x (16.00 g/mol) = 108.01 g/mol
Next, look at the balanced chemical equation:
2N2O5(s) → 4NO2(g) + O2(g)
From the equation, you can see that for every 2 moles of N2O5, 1 mole of O2 is produced. Therefore, the molar ratio of N2O5 to O2 is 2:1.
Now, use the given mass of N2O5 (44.2 mg) and convert it to moles:
44.2 mg ÷ 108.01 g/mol = 0.4093 mmol (millimoles)
Since the molar ratio of N2O5 to O2 is 2:1, the moles of O2 formed will be half of the moles of N2O5:
0.4093 mmol ÷ 2 = 0.2047 mmol
Finally, convert the moles of O2 back to grams using its molar mass:
0.2047 mmol x 32.00 g/mol = 6.55 mg
Therefore, from 44.2 mg of N2O5, approximately 6.55 mg of oxygen (O2) will be formed.