n_P=m/M={4,65}/31=0,15(mol)

n_{O_2}=V/{24,79}={3,7185}/{24,79}=0,15(mol)

a, 4P+5O_2 $$\xrightarrow{t^o}$$ 2P_2O_5

b, SS: n_P > n_{O_2}  ({0,15}/4 > {0,15}/5)

-> P\ dư.

Theo\ pt:n_P\ p//ư=4/5xxn_{O_2}=4/5 xx 0,15=0,12(mol)

-> n_P\ dư=n_{bđ}-n_{p//ư}=0,15-0,12=0,03(mol)

-> m_P\ dư=n xx M=0,03 xx 31=0,93(g)

c,

Theo\ pt:n_{P_2O_5}=2/5 xx n_{O_2}=0,06(mol)

-> m_{P_2O_5}=n xx M=0,06 xx 142=8,52(g)