According to the following reaction, how many grams of nitrogen gas will be formed upon the complete reaction of 22.2 grams of hydrogen gas with excess nitrogen monoxide?
nitrogen monoxide (g) + hydrogen (g) nitrogen (g) + water (l)
You need to write and balance the equation.
2NO + 2H2 ==> N2 + 2H2O
Convert 22.2 g H2 to mols. mols = grams/molar mass.
Use the coefficients in the balanced equation to convert mols H2 to mols N2.
Now convert mols N2 gas to grams. g = mols x molar mass.
To determine the number of grams of nitrogen gas formed, we need to first balance the chemical equation:
2 NO (g) + 2 H2 (g) -> N2 (g) + 2 H2O (l)
Now, we can see that for every 2 moles of nitrogen monoxide (NO) and 2 moles of hydrogen (H2) reacted, we get 1 mole of nitrogen (N2) formed.
Next, we need to convert the given mass of hydrogen gas (H2) to moles. We can use the molar mass of hydrogen, which is approximately 2 grams/mol, to calculate this conversion.
22.2 grams H2 * (1 mol H2 / 2 g H2) = 11.1 mol H2
Since the reaction is in a 1:1 ratio for moles of hydrogen and nitrogen, we can conclude that 11.1 moles of nitrogen gas will be formed.
Lastly, we can convert moles of nitrogen gas to grams using its molar mass. The molar mass of nitrogen (N2) is approximately 28 grams/mol.
11.1 mol N2 * (28 g N2 / 1 mol N2) = 310.8 grams N2
Therefore, 310.8 grams of nitrogen gas will be formed upon the complete reaction.
To answer this question, we need to use stoichiometry to calculate the mass of nitrogen gas formed.
First, the balanced chemical equation is:
2NO (g) + 2H2 (g) → N2 (g) + 2H2O (l)
From the balanced equation, we can see that it requires 2 moles of NO to produce 1 mole of N2.
To find the number of moles of NO in 22.2 grams, we need to use the molar mass of NO, which is 30.01 g/mol.
Number of moles of NO = Mass of NO / Molar mass of NO
= 22.2 g / 30.01 g/mol
= 0.739 moles
According to the stoichiometry of the balanced equation, we know that 2 moles of NO produce 1 mole of N2. Therefore, the number of moles of N2 formed will be half of the moles of NO.
Number of moles of N2 = 0.739 moles / 2
= 0.3695 moles
Finally, to calculate the mass of N2 formed, we multiply the number of moles by the molar mass of N2, which is 28.01 g/mol.
Mass of N2 = Number of moles of N2 × Molar mass of N2
= 0.3695 moles × 28.01 g/mol
= 10.34 grams
Therefore, 10.34 grams of nitrogen gas will be formed upon the complete reaction of 22.2 grams of hydrogen gas with excess nitrogen monoxide.