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.
To solve this problem, we need to use stoichiometry to determine the number of moles of K2PtCl4 that will react to produce 64.1 g of PtCl2(NH3)2.
First, we need to find the molar mass of PtCl2(NH3)2.
The molar mass of Pt is 195.08 g/mol.
The molar mass of Cl is 35.45 g/mol.
The molar mass of N is 14.01 g/mol.
The molar mass of H is 1.01 g/mol.
PtCl2(NH3)2 contains 1 Pt atom, 2 Cl atoms, and 4 NH3 molecules, so the total molar mass is:
(1 x Molar mass of Pt) + (2 x Molar mass of Cl) + (4 x (Molar mass of N + Molar mass of H))
= (1 x 195.08 g/mol) + (2 x 35.45 g/mol) + (4 x (14.01 g/mol + 1.01 g/mol))
= 195.08 g/mol + 70.90 g/mol + 60.48 g/mol
= 326.46 g/mol
Next, we can use the balanced chemical equation to determine the stoichiometric ratio between K2PtCl4 and PtCl2(NH3)2.
According to the equation, 1 mol of K2PtCl4 reacts to produce 1 mol of PtCl2(NH3)2.
Now we can calculate the number of moles of PtCl2(NH3)2 produced:
Number of moles = Mass / Molar mass
Number of moles = 64.1 g / 326.46 g/mol
Number of moles ≈ 0.1965 mol
Therefore, approximately 0.1965 mol of K2PtCl4 must react to produce 64.1 g of PtCl2(NH3)2.
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