Решите задачу по химии 9 класс

Mr (Al2S3)=27*2+32*3=54+96=150 г/моль W(Al) = 2[latex] \frac{Ar(Al)*k}{Mr(Al2S3)*100} = \frac{27*2}{150}*100=36% [/latex] W(S) = [latex] \frac{32*3}{150}*100=64% [/latex] ω(Al) = 2[latex] \frac{Ar(Al)*k}{Mr(Al2S3)*100} = \frac{27*2}{150}*100=36% [/latex] ω(S) = [latex] \frac{32*3}{150}*100=64% [/latex] 2m(Al) : 3m(S) = 2*27 : 3*32 = 54 : 96