Obilecz ile gramów stanowi 1/5 mola CO/// 0,25 mola Mg(NO3)2/// 0,75 mola Ca3(PO4)2
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Obilecz ile gramów stanowi 1/5 mola CO/// 0,25 mola Mg(NO3)2/// 0,75 mola Ca3(PO4)2
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1)
M CO=12+16=28
ms=M*n
ms=5.6g
2)
M Mg(NO3)2=24+2*(14+3*16)=24+2*62=148
ms=37g
3)
M Ca3(PO4)2=40*3+2*(31+16*4)=120+190=310
ms=232.5g