How much K2PtCl4 must react in order to produce 74g of PtCl2(NH3)2

PART 1)
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) + 2NH3(aq) -----> 2KCl(aq) + PtCl2(NH3)2(aq)
How much K2PtCl4 must react in order to produce 74g of PtCl2(NH3)2? Answer in units of mol.

PART 2)
How much NH3 is needed to produce 74g of PtCl2(NH3)2? Answer in units of mol please.

If you can answer these thanks so much!!!

K2PtCl4(aq) + 2NH3(aq) -----> 2KCl(aq) + PtCl2(NH3)2(aq)

74g PtCl2(NH3)2 / 300g/mole = 0.247moles PtCl2(NH3)2
0.247moles (1K2PtCl4 / 1PtCl2(NH3)2) = 0.247moles K2PtCl4 required

NH3:PtCl2(NH3)2 = 2:1 so 0.493moles NH3 required