HOW MANY LITERS OF H2 must react to produce 75.0g of PH3
To determine the number of liters of H2 needed to produce 75.0 grams of PH3, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between H2 and PH3:
2H2 + P4 → 4PH3
Step 2: Calculate the molar mass of PH3:
The molar mass of PH3 can be calculated as follows:
(1 atom of P x atomic mass of P) + (3 atoms of H x atomic mass of H)
= (1 x 31.0 g/mol) + (3 x 1.0 g/mol)
= 31.0 g/mol + 3.0 g/mol
= 34.0 g/mol
Step 3: Convert the given mass of PH3 to moles:
To do this, divide the mass (75.0 g) by the molar mass (34.0 g/mol):
75.0 g PH3 / 34.0 g/mol = 2.21 mol PH3
Step 4: Use the mole ratio to find the number of moles of H2 required:
From the balanced equation, we see that for every 4 moles of PH3 produced, 2 moles of H2 are required. Therefore, the mole ratio of H2 to PH3 is 2:4 or 1:2.
Since we have 2.21 moles of PH3, we need twice as many moles of H2:
2.21 mol PH3 x (2 mol H2 / 4 mol PH3) = 1.11 mol H2
Step 5: Convert moles of H2 to liters:
To convert from moles to liters of a gas, we need to apply the ideal gas law (PV = nRT).
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
Therefore, the number of liters of H2 can be calculated as follows:
1.11 mol H2 x (22.4 L / 1 mol H2) = 24.8 L H2
So, 24.8 liters of H2 are needed to produce 75.0 grams of PH3.
To determine the number of liters of H2 gas required to produce 75.0 grams of PH3, we need to apply stoichiometry calculations using balanced chemical equations.
Let's start with the balanced equation for the reaction:
4 H2 + P4 -> 4 PH3
We can see from the equation that it takes 4 moles of H2 to produce 4 moles of PH3.
To find the number of moles of PH3 from the given mass, we need to divide the given mass by the molar mass of PH3.
The molar mass of PH3 (phosphine) is calculated as follows:
P = 1 * atomic mass of P = 1 * 31.0 g/mol = 31.0 g/mol
H = 3 * atomic mass of H = 3 * 1.0 g/mol = 3.0 g/mol
Molar mass of PH3 = 31.0 g/mol + 3.0 g/mol = 34.0 g/mol
Now, we can calculate the number of moles of PH3 by dividing the given mass by the molar mass:
Number of moles of PH3 = 75.0 g / 34.0 g/mol ≈ 2.21 mol
Since the reaction is 4 H2 + P4 -> 4 PH3, we know that 4 moles of H2 react to produce 4 moles of PH3. Therefore, the number of moles of H2 required to produce 2.21 moles of PH3 is also 2.21 moles.
Now, we can use the ideal gas law to convert moles of H2 to liters of H2.
The ideal gas law is given by:
PV = nRT
Assuming standard temperature and pressure (STP), the values are as follows:
P = 1 atm (pressure)
V = unknown (volume in liters)
n = 2.21 mol (number of moles of H2)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 273.15 K (temperature at STP)
Rearranging the equation to solve for V, we have:
V = (n * R * T) / P
Substituting the values, we get:
V = (2.21 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm
Simplifying the equation, we find:
V ≈ 50.4 L
Therefore, approximately 50.4 liters of H2 gas are required to produce 75.0 grams of PH3.
Here is a solved example of this type problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html