How much phosphine PH3 will be released when 70 g of calcium phosphide Ca3P2 interacts with hydrochloric acid?
| December 3, 2020 | education
How much phosphine PH3 will be released when 70 g of calcium phosphide Ca3P2 interacts with hydrochloric acid? How much hydrochloric acid is required. Ca3P2 + 6HCI → 3CaCI2 + 2PH3
Ca3P2 + 6HCI → 3CaCI2 + 2PH3
n (Ca3P2) = m \ M = 70 \ 182 = 0.38 mol
n (PH3) = 0.38 * 2 = 0.76 mol according to the reaction equation
V (PH3) = n * Vm = 0.76 * 22.4 = 17.024 l
n (HCl) = 0.38 * 6 = 2.28 mol according to the reaction equation
Answer: n (HCl) = 2.28 mol
V (PH3) = 17.024 l
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
