how many grams nitrogen dioxide are formed if 0.46 mol of ammonium nitrate is used
To determine the number of grams of nitrogen dioxide formed when 0.46 mol of ammonium nitrate is used, we need to utilize the balanced chemical equation for the reaction. The balanced equation for the decomposition reaction of ammonium nitrate is as follows:
NH4NO3 → N2O + 2H2O
From this equation, we can see that 1 mol of ammonium nitrate produces 1 mol of nitrogen dioxide (N2O). Thus, the number of moles of nitrogen dioxide formed can be calculated as:
0.46 mol (NH4NO3) × 1 mol (N2O) / 1 mol (NH4NO3) = 0.46 mol (N2O)
To convert this to grams, we need to multiply the number of moles of nitrogen dioxide by its molar mass. The molar mass of N2O is:
2 × atomic mass of nitrogen (N) + atomic mass of oxygen (O)
= 2 × 14.01 g/mol + 16.00 g/mol
= 28.02 g/mol + 16.00 g/mol
= 44.02 g/mol
Now, we can calculate the mass of nitrogen dioxide formed:
Mass (N2O) = 0.46 mol (N2O) × 44.02 g/mol = 20.19 grams
Therefore, if 0.46 mol of ammonium nitrate is used, approximately 20.19 grams of nitrogen dioxide will be formed.