Answer to Question #108966 in General Chemistry for alex

Question

2.5g of P4 reacts with an excess amount of F2 according to the reaction shown below. determine the maximum amount of PF5 produced? (P: 30.97g/mol, F:19g/mol)
P4+10F2 -> 4PF5

Answer 1

Solution:

P₄+10F₂ -> 4PF₅

n_PF5=4n_P4 - according to stoichiometric ratios and with an excess of fluorine

n_P4=m_P4/M_P4=2.5g/(4*30.97g*mol^-1)=0.02mol

n_PF5=4*0.02mol=0.08mol

m_PF5=n_PF5*M_PF5=0.08mol*(30.97g*mol^-1+5*19g*mol^-1)=10.1g

Answer:

Maximum amount of PF5 - 10,1g