请教一道化学计算大题目

匿名用户 | 2017-05-23 09:05

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  • 025)所以m=2;3,S.2840mgCa.00400m+0;n[Ca(H2PO4)2]=x所以n(H3PO4)=200x/,P的物质的量分别为0,0;2n(P)所以0.00044mmol,0,P的物质的量分别为0;(x+2)n(Ca(H2PO4)2)=200/.0350mg.00044m+0;2(0.0350m/:P<56=0.4730mg.00845m=0;80=0.912)注意H3PO4和Ca(H2PO4)2中Ca,0:2所以应该利用磷来计算n(H3PO4)+2n(Ca(H2PO4)2)=100*2=200而n[H3PO4]/.2840m/.4730m/.00400mmol混酸中S.005mol最后得到的只有CaSO4和Ca(H2PO4)2n(Ca)=n(S)+1/,0.0005+1/,01)磷灰石粉末中CaO,SO3;71=0.025mol;(x+2)所以y=a*n[H3PO4]+b*n[Ca(H2PO4)=(200ax+200b)/,P2O5的质量分别为0.00845mmol
    匿名用户 | 2017-05-23 09:05

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