Reacting Masses No. moles = mass M r n = m M r
What mass of ammonium nitrate fertiliser can be made from 18.9 g of nitric acid? NH 3 + HNO 3 → NH 4 NO 3 1 mole → 1 mole 1. Write balanced equation 3. Calc n o. moles HNO 3 = mass / M r N o moles HNO 3 = = 0.3 mol 2. Calc M r of HNO 3 M r = (3x16) = 63
What mass of ammionium nitrate fertiliser can be made from 18.9 g of nitric acid? No. moles NH 4 NO 3 = 0.3 mol 5. Calculate mass of NH 4 NO 3 = No. moles x Mr Mass NH 4 NO 3 = 0.3 x 80= 24g 4. Calc n o. moles NH 4 NO 3 from equation 1 mole → 1 mole NH 3 + HNO 3 → NH 4 NO 3 M r = (3 x 16) = 80
A farmer requires 33kg of ammonium sulphate fertiliser. How much ammonia is needed ? 2NH 3 + H 2 SO 4 → (NH 4 ) 2 SO 4 2 mole → 1 mole 1. Write balanced equation 2. Calc M r of (NH 4 ) 2 SO 4 N o moles (NH 4 ) 2 SO 4 = = 0.25 mol 3. Calc n o. moles (NH 4 ) 2 SO 4 = mass / M r 2 x (14 + 4) (4 x 16) = 132
A farmer requires 33kg of ammonium sulphate fertiliser. How much NH 3 is needed ? No. moles NH 3 = 2 x mol (NH 4 ) 2 SO 4 5. Calculate mass of NH 3 = No. moles x Mr Mass NH 3 = 0.5 x 17 = 8.5g 4. Calc n o. moles NH 3 from equation 2 mole → 1 mole 2NH 3 + H 2 SO 4 → (NH 4 ) 2 SO 4 No. moles NH 3 = 2 x 0.25= 0.5