K2PtCl4 + NH3 = 2 KCl + PtCl2(NH3) the mass of cisplatin produced from 2.5 g K2PtCl4 reacting with excess ammonia
To calculate the mass of cisplatin produced, we first need to determine the limiting reagent in the reaction between K2PtCl4 and NH3. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.
Let's start by calculating the molar masses of the reactants involved:
- Molar mass of K2PtCl4 = 581.1 g/mol (From periodic table)
- Molar mass of NH3 = 17.03 g/mol (From periodic table)
Next, we can calculate the number of moles of K2PtCl4 in 2.5 g:
Number of moles of K2PtCl4 = Mass / Molar mass
= 2.5 g / 581.1 g/mol
≈ 0.0043 mol
Since K2PtCl4 has a stoichiometric coefficient of 1 in the balanced equation, it means that 0.0043 moles of K2PtCl4 will react to form an equal amount of moles of cisplatin (PtCl2(NH3)).
The balanced equation tells us that 1 mol of K2PtCl4 produces 1 mol of cisplatin. Hence, we can say that 0.0043 mol of K2PtCl4 will produce 0.0043 mol of cisplatin.
Now, we need to convert the moles of cisplatin to grams. To do this, we can use the molar mass of cisplatin (PtCl2(NH3)).
Molar mass of PtCl2(NH3) = Molar mass of Pt + 2 * Molar mass of Cl + Molar mass of N + 3 * Molar mass of H
≈ 195.1 g/mol + 2 * 35.45 g/mol + 14.01 g/mol + 3 * 1.01 g/mol
≈ 300.135 g/mol
To calculate the mass of cisplatin produced, we can multiply the moles of cisplatin by its molar mass:
Mass of cisplatin = number of moles * molar mass
= 0.0043 mol * 300.135 g/mol
≈ 1.29 g
Therefore, approximately 1.29 grams of cisplatin will be produced when 2.5 grams of K2PtCl4 reacts with excess ammonia.
convert to moles
Each mole of K2PtCl4 produces one mole of PtCl2(NH3)
convert back to grams