First figure out the limiting reactant by calculating how many moles of product you get from each starting material.
For the 55.0g of K2PtCl4, convert that to moles by dividing by the molar mass (415.1 g/mol) to get .132 moles K2PtCl4. You get 1 mole of product for ever mole of K2PtCl4, so you'd also get .132 moles of the product.
For the 35.6 g of NH3, convert to moles (MM=17g/mol) to get 2.09 moles NH3. It takes two moles of NH3 to get one mole of product, so you'd be able to produce 1.05 moles of product from that.
K2PtCl4 is the limiting reactant, because you only have enough of it to make .132 moles of product, and when its all used up, it doesn't matter how much NH3 you have, you won't get any more product.
So we can form .123 moles of product with these conditions.
Convert that to grams by multiplying by the molar mass (300 g/mol) to get 36.9 g product. That's how much you'd get if the reaction was 100% efficient. You'd only get 95% of that, however, so 35.1g.
