At STP, what is the volume of HBr that will result from the reaction of 10 liters of H2 and 10 liters of Br2?
Work would be nice :]
H2 + Br2 ==> 2HBr.
When all are gases, one can use volume directly and not go through converting to moles.
10 L H2 x (2 moles HBr/1 mole H2) = 20 L HBr.
10 L Br2 x (2 moles HBr/1 mole Br2) = 20 L HBr.
So 20 L HBr is the product.
To determine the volume of HBr produced from the reaction of 10 liters of H2 and 10 liters of Br2 at STP (Standard Temperature and Pressure), we need to use the stoichiometry of the balanced chemical equation and the ideal gas law.
First, let's start by writing the balanced chemical equation for the reaction between H2 and Br2 to form HBr:
H2 + Br2 -> 2HBr
From the balanced equation, we can see that 1 mole of H2 reacts with 1 mole of Br2 to produce 2 moles of HBr.
Next, let's convert the given volumes of H2 and Br2 into moles using the ideal gas law equation, PV = nRT. At STP, the pressure (P) is 1 atm, the temperature (T) is 273.15 K, and the ideal gas constant (R) is 0.0821 L·atm/(mol·K).
For H2:
P = 1 atm
V = 10 L
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
Using PV = nRT, we can rearrange the equation to solve for n (moles):
n(H2) = (P * V) / (R * T)
n(H2) = (1 atm * 10 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
= 0.445 mol
Similarly, we can calculate the number of moles of Br2 using the same equation. Since the stoichiometry between H2 and Br2 is 1:1, the number of moles of Br2 is also 0.445 mol.
Now, let's use the stoichiometry of the equation to determine the number of moles of HBr produced. From the balanced equation, we know that 1 mole of H2 produces 2 moles of HBr:
n(HBr) = 2 * n(H2)
= 2 * 0.445 mol
= 0.89 mol
Finally, let's convert the moles of HBr into volume using the ideal gas law:
V(HBr) = (n * R * T) / P
V(HBr) = (0.89 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
= 20.34 L
Therefore, at STP, the volume of HBr that will result from the reaction of 10 liters of H2 and 10 liters of Br2 is approximately 20.34 liters.