The compound PtCl2 (NH3)2 is effective as a treatment for some cancers. It is synthesized by the reaction shown in the following equation. K2 PtCl4(aq) + 2NH3 (aq) ⟶ 2KCl (aq) + PtCl2 (NH3)2(aq). How many moles of K2 PtCl4 must react in order to produce 30.0 g PtCl2 (NH3)2?A) 0.100 mol K2 PtCl4B) 0.200 mol K2 PtCl4C) 0.140 mol K2 PtCl4D) 0.160 mol K2 PtCl4

Answer:
A) 0.100 mol K2 PtCl4

Explanation:
From the periodic table:
atomic mass of Pt = 195 grams
atomic massof Cl = 35.5 grams
atomic mass of N = 14 grams
atomic mass of H = 1 gram
Therefore:
molar mass of PtCl2 (NH3)2 = 195+2(35.5)+2(14)+6(1) = 300 grams/mol
Now, we know that mass of PtCl2 (NH3)2 produced is 30 grams. We will get the number of moles of the produced PtCl2 (NH3)2 as follows:
number of moles = mass / molar mass
number of moles = 30 / 300 = 0.1 mole

Now, from the balanced equation given:
1 mole of K2 PtCl4 produces 1 mole of PtCl2 (NH3)2.
Therefore, the ratio between the two is 1:1 which means that 0.1 mole of PtCl2 (NH3)2 will be produced from 0.1 mole of K2 PtCl4

Hope this helps :)