If .237 mol of ammonia react with 56g of iodine, how much Nitrogen triiodid, in grams, is produced?
To determine the amount of nitrogen triiodide (NI3) produced, we need to first write and balance the chemical equation for the reaction between ammonia (NH3) and iodine (I2):
NH3 + I2 → NI3 + HI
The balanced equation indicates that 1 mole of ammonia reacts with 1 mole of iodine to produce 1 mole of nitrogen triiodide.
Given that 0.237 mol of ammonia reacts, we can use the molar ratio to determine the moles of nitrogen triiodide produced. Since the ratio is 1:1, we have 0.237 mol of nitrogen triiodide.
To calculate the mass of nitrogen triiodide produced, we need the molar mass of NI3. Adding together the atomic masses of nitrogen (N) and three iodine (I) atoms, we have:
Molar mass of NI3 = (1 × atomic mass of N) + (3 × atomic mass of I)
Molar mass of NI3 = (1 × 14.01 g/mol) + (3 × 126.90 g/mol)
Molar mass of NI3 = 14.01 g/mol + 380.70 g/mol
Molar mass of NI3 = 394.71 g/mol
Now, to calculate the mass of NI3 produced, we can use the following formula:
Mass = moles × molar mass
Mass of NI3 produced = 0.237 mol × 394.71 g/mol
Mass of NI3 produced = 93.56 g
Therefore, approximately 93.56 grams of nitrogen triiodide is produced.