Question

某硝酸铵化肥中含氮31.5%,求该化肥的纯度?

Answer 1

解

设 硝酸铵化肥纯度为x,质量为m

NH4NO3 中 N 的质量分数 = 2N/NH4NO3 = 28/80 = 35%

m * x * 35% = m * 31.5%

x = 90%

答：硝酸铵化肥纯度为90%.