64.1 g of PtCl2(NH3)2? change that to moles! I think it is about 1/5 of a mole (https://www.webqc.org/molecular-weight-of-PtCl2%28NH3%292.html)
So the balanced equation indicates the same number of moles of K2PtCl4
The compound PtCl2(NH3)2 is effective as a
treatment for some cancers. It is synthesized
by the reaction shown in the equation
K2PtCl4(aq) + 2 NH3(aq) →
2 KCl(aq) + PtCl2(NH3)2(aq).
How much K2PtCl4 must react in order to
produce 64.1 g of PtCl2(NH3)2?
Answer in units of mol.
1 answer