Question

, нужно рассчитать массовые доли каждого элемента : 1. NaOH 2.Mg(NO3)2

Answer 1

№1.$m_{r}(NaOH) = A_{r} (Na) + A_{r} (O) + A_{r} (H) = 23+16+1 = 40.$ω(Na) = $\frac{1*A_{r} (Na)}{m_{r} (NaOH)} =\frac{23}{40} = 0,575 = 57,5 \%.$ω(О) = $\frac{1*A_{r} (O)}{m_{r} (NaOH)} =\frac{16}{40} = 0,4 = 40 \%.$ω(H) = $\frac{1*A_{r} (H)}{m_{r} (NaOH)} =\frac{1}{40} = 0,025 = 2,5 \%.$

ответ : ω(Na) = 57,5%, ω(O) = 40%, ω(H) = 2,5%.

№2. $m_{r} (Mg(NO_{3})_{2}) = A_{r}(Mg)+ 2*A_{r} (N)+6*A_{r} (O) = 24+2*14+6*16 = 24+28+96=148.$ω(Mg) = $\frac{1*A_{r}(Mg) }{m_{r} (Mg(NO_{3})_{2})} = \frac{24}{148}$ ≈ 0,162 ≈ 16,2%.ω(N) = $\frac{2*A_{r}(N) }{m_{r} (Mg(NO_{3})_{2})} =\frac{28}{148}$ ≈ 0,189 ≈ 18,9%.ω(O) = $\frac{6*A_{r}(O) }{m_{r} (Mg(NO_{3})_{2})} = \frac{96}{148}$ ≈ 0,649 ≈ 64,9%.

ответ : ω(Mg) ≈ 16,2%, ω(N) ≈ 18,9%, ω(O) ≈ 64,9%.