1. Cisplatin is an anti tumor agent. It has the molecular formula Pt(NH3)2Cl2. How many grams of cisplatin can be produced if the limiting reactant is 1 kg of platinum?

I got 1538.1 g Pt(NH3)2Cl2.

2. Hydrogen cyanide is used in the production of cyanimid fertilizers. It is produced by the following reaction.
2 CH4+2NH3+3O2---> 2HCN+6H2O

A. How much HCN can be produced starting with 100 kg of each of the reactants?

Is it 0.0562 g?

B. What is the limiting reactant?

Oxygen?

1 is ok with the number but you aren't allowed that many significant figures.

2. No. I think you came out with the right answer for kg (56.2 kg) and converted to grams wrong.

To find the answer to the first question, we need to calculate the molar mass of cisplatin and then determine the number of moles of platinum in 1 kg. Finally, we can use stoichiometry to find the number of moles of cisplatin produced and convert it to grams.

1. The molecular formula of cisplatin is Pt(NH3)2Cl2. The molar mass can be calculated by summing up the atomic masses of each element in the formula.

Pt: 195.08 g/mol
N: 14.01 g/mol
H: 1.01 g/mol
Cl: 35.45 g/mol

Molar mass of cisplatin = (1 * 195.08) + (2 * 14.01) + (6 * 1.01) + (2 * 35.45) = 300.05 g/mol

2. Now, we need to determine the number of moles of platinum in 1 kg. The molar mass of platinum is 195.08 g/mol.

Number of moles of platinum = (1,000 g) / (195.08 g/mol) = 5.126 mol

3. According to the balanced equation, the stoichiometric ratio between platinum and cisplatin is 1:1. Therefore, the number of moles of cisplatin produced will be the same as the number of moles of platinum.

Number of moles of cisplatin = 5.126 mol

4. Finally, we can convert the number of moles of cisplatin to grams by multiplying it by the molar mass of cisplatin.

Mass of cisplatin = (5.126 mol) * (300.05 g/mol) = 1537.85 g

Therefore, the mass of cisplatin that can be produced from 1 kg of platinum is approximately 1537.85 grams.

For the second question:

A. To find the amount of HCN produced, we need to use stoichiometry and consider the limiting reactant. In this case, we start with 100 kg of each reactant, so we need to determine the limiting reactant.

First, let's calculate the number of moles for each reactant:

Number of moles of CH4 = (100,000 g) / (16.04 g/mol) = 6245.11 mol
Number of moles of NH3 = (100,000 g) / (17.03 g/mol) = 5876.52 mol
Number of moles of O2 = (100,000 g) / (32.00 g/mol) = 3125.00 mol

The stoichiometric ratio between CH4, NH3, and O2 is 2:2:3. Therefore, the limiting reactant will be the one that produces the smallest amount of HCN.

Using the stoichiometry, we can calculate the number of moles of HCN that will be produced if each reactant is completely consumed:

Number of moles of HCN from CH4 = (6245.11 mol CH4) * (2 mol HCN / 2 mol CH4) = 6245.11 mol
Number of moles of HCN from NH3 = (5876.52 mol NH3) * (2 mol HCN / 2 mol NH3) = 5876.52 mol
Number of moles of HCN from O2 = (3125.00 mol O2) * (2 mol HCN / 3 mol O2) = 2083.33 mol

Since the smallest value is 2083.33 mol, we can conclude that the number of moles of HCN produced is 2083.33 mol.

To convert this to grams, we need to multiply by the molar mass of HCN:

Mass of HCN = (2083.33 mol) * (27.03 g/mol) = 56376.67 g ≈ 56.37 kg

Therefore, the amount of HCN that can be produced is approximately 56.37 kg.

B. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. In this case, the limiting reactant is the reactant that produces the smallest amount of product.

From the stoichiometry calculations above, we found that the number of moles of HCN produced from CH4 is 6245.11 mol, from NH3 is 5876.52 mol, and from O2 is 2083.33 mol.

Since the smallest value is 2083.33 mol, we can conclude that the limiting reactant is oxygen (O2).