Nickel carbonyl has a formula: Ni(CO)4. Nickel carbonyl is produced by the reaction of solid nickel metal with carbon monoxide gas.
1. how do you write a balanced equation for this reaction?
2. How many liters of carbon monoxide gas must be provided to react with 5.00g of nickel metal at STP?
3. How many grams of nickel carbonyl would be formed by the reaction of #2?
Ni(s) + 4CO(g) ==> Ni(CO)4(l)
The other two parts are stoichiometry problems. Here is a link that gives secific instructions for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
1. To write a balanced equation for the reaction between solid nickel metal (Ni) and carbon monoxide gas (CO), we need to ensure that the number of atoms of each element is the same on both sides of the equation.
The balanced equation for the reaction can be written as follows:
Ni + 4CO -> Ni(CO)4
2. To calculate the number of liters of carbon monoxide gas required to react with 5.00g of nickel metal at standard temperature and pressure (STP), we need to use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
At STP, the pressure is 1 atm and the temperature is 273K. The ideal gas constant is approximately 0.0821 L.atm/(mol.K).
First, we need to convert the mass of nickel (5.00g) to moles. The molar mass of nickel is 58.6934 g/mol.
moles of Ni = mass of Ni / molar mass of Ni
moles of Ni = 5.00g / 58.6934 g/mol
moles of Ni ≈ 0.0853 mol
According to the balanced equation, the stoichiometric ratio between Ni and CO is 1:4. Therefore, we need 4 times as many moles of CO as moles of Ni.
moles of CO = 4 * moles of Ni
moles of CO = 4 * 0.0853 mol
moles of CO ≈ 0.3412 mol
Now, we can use the ideal gas law equation to calculate the volume of carbon monoxide gas.
V = (n * R * T) / P
V = (0.3412 mol * 0.0821 L.atm/(mol.K) * 273K) / 1 atm
V ≈ 7.45 L
Therefore, approximately 7.45 liters of carbon monoxide gas are required.
3. To determine the number of grams of nickel carbonyl (Ni(CO)4) formed by the reaction calculated in question 2, we need to use the molar mass of the compound.
The molar mass of Ni(CO)4 can be calculated as follows:
molar mass of Ni = 58.6934 g/mol
molar mass of CO = (12.0107 g/mol * 1) + (15.9994 g/mol * 1) = 28.0101 g/mol
molar mass of Ni(CO)4 = (58.6934 g/mol) + (4 * 28.0101 g/mol)
molar mass of Ni(CO)4 ≈ 170.2358 g/mol
Using the stoichiometry of the balanced equation, we know that the ratio between the moles of Ni and the moles of Ni(CO)4 is 1:1.
moles of Ni(CO)4 = moles of Ni
moles of Ni(CO)4 ≈ 0.0853 mol
Now we can calculate the mass of nickel carbonyl:
mass of Ni(CO)4 = moles of Ni(CO)4 * molar mass of Ni(CO)4
mass of Ni(CO)4 ≈ 0.0853 mol * 170.2358 g/mol
mass of Ni(CO)4 ≈ 14.545 g
Therefore, approximately 14.545 grams of nickel carbonyl would be formed by the reaction.
1. To write a balanced equation for the reaction between solid nickel metal and carbon monoxide gas, you need to ensure that the number of atoms of each element is the same on both sides of the equation.
The reaction can be written as follows:
Ni + 4CO → Ni(CO)4
By balancing the equation, you need to make sure there is an equal number of atoms for each element on both sides of the arrow. In this case, there is 1 nickel (Ni) atom and 4 carbon monoxide (CO) molecules on the left side and 1 nickel carbonyl (Ni(CO)4) molecule on the right side.
2. To determine the amount of carbon monoxide gas required to react with 5.00g of nickel metal at STP (Standard Temperature and Pressure), you need to use the ideal gas law and stoichiometry.
First, you need to convert the mass of nickel metal to the number of moles using its molar mass. The molar mass of nickel (Ni) is 58.69 g/mol. Divide the given mass by the molar mass:
5.00 g Ni / 58.69 g/mol = 0.0852 mol Ni
According to the balanced equation, it requires a 1:4 ratio of nickel (Ni) to carbon monoxide (CO). Therefore, for every 1 mole of nickel, you will need 4 moles of carbon monoxide.
0.0852 mol Ni * 4 mol CO / 1 mol Ni = 0.3408 mol CO
Now, you need to convert the moles of carbon monoxide to Liters using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
At STP, the temperature is 273.15 K, and the pressure is 1 atmosphere. Plug in the values:
V = (0.3408 mol CO * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
V ≈ 7.49 L
Therefore, approximately 7.49 Liters of carbon monoxide gas must be provided to react with 5.00g of nickel metal at STP.
3. To determine the grams of nickel carbonyl (Ni(CO)4) formed by the reaction in question 2, you need to use the stoichiometry of the balanced equation.
From question 2, we know that 0.3408 mol of CO is required. According to the balanced equation, there is a 1:1 ratio between the moles of carbon monoxide (CO) and nickel carbonyl (Ni(CO)4).
Therefore, the number of moles of nickel carbonyl formed is also 0.3408 mol.
To convert the moles of nickel carbonyl to grams, you need to multiply the number of moles by its molar mass. The molar mass of nickel carbonyl (Ni(CO)4) can be calculated by adding the molar masses of the individual elements:
Molar mass Ni(CO)4 = Molar mass Ni + (4 x Molar mass C) + (4 x Molar mass O)
Molar mass Ni(CO)4 = 58.69 g/mol + (4 x 12.01 g/mol) + (4 x 16.00 g/mol) ≈ 170.73 g/mol
Now, multiply the number of moles by the molar mass:
0.3408 mol Ni(CO)4 * 170.73 g/mol = 58.23 g
Therefore, approximately 58.23 grams of nickel carbonyl would be formed by the reaction in question 2.