hydrogen gas can be produced by reacting aluminum with sulfuric acid .How many L hydrogen gas are produced by reacting 15.0 mol of sulfuric acid with 15.0 mol of aluminum at STP?
This is a limiting reagent problem. I now that because amounts are given for BOTH reactants. Write a balanced equation.
2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2
Convert mols of each reactant to mols H2.
Al first.
15.0 mols Al x [3 mols H2/2 mols Al) = 15.0 x 3/2 = 22.5 mols H2 gas.
Next H2SO4:
15.0 mols H2SO4 x (3 mols H2/3 mols H2SO4) = 15.0 x 1/1 = 15.0 mol H2 produced.
The two answers are different; therefore, one of them must be incorrect. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. H2SO4 is the LR.
1 mol gas at STP occupies a volume of 22.4L; therefore, 15.0 mols x 22.4L = ? L H2 gas from this reaction if you have 100% yield.
what would the volume of the gas produced be at a pressure of 700mmHg and a temperature of 30degrees of C
Use PV = nRT
You have n.Substitute the new T and P. Remember T must be in kelvin and P in atmospheres.
P of 700 mm = 700/760 atm
To determine the volume of hydrogen gas produced by reacting 15.0 moles of sulfuric acid (H2SO4) with 15.0 moles of aluminum (Al) at STP (Standard Temperature and Pressure), we can use the balanced chemical equation for the reaction:
2Al + 3H2SO4 → 3H2 + Al2(SO4)3
From the balanced equation, we can see that 3 moles of hydrogen gas (H2) are produced for every 2 moles of aluminum (Al). Therefore, we can use the mole ratio to calculate the moles of hydrogen gas produced.
Given:
Moles of sulfuric acid (H2SO4) = 15.0 mol
Moles of aluminum (Al) = 15.0 mol
Since we have the same number of moles for both reactants, we can determine which reactant is limiting by comparing the mole ratio from the balanced equation.
From the balanced equation:
2 moles of aluminum (Al) produce 3 moles of hydrogen gas (H2)
Since 2 moles of aluminum produce a greater number of moles of hydrogen gas than 15.0 moles of sulfuric acid, we can conclude that the sulfuric acid is the limiting reactant.
To calculate the moles of hydrogen gas produced, we can use the mole ratio:
Moles of hydrogen gas (H2) = Moles of sulfuric acid (H2SO4) × (3 moles of H2 / 3 moles of H2SO4)
Moles of hydrogen gas (H2) = 15.0 mol × (3 mol of H2 / 3 mol of H2SO4) = 15.0 mol
Therefore, 15.0 moles of hydrogen gas (H2) would be produced when reacting 15.0 moles of sulfuric acid (H2SO4) with 15.0 moles of aluminum (Al) at STP.
To convert the moles of hydrogen gas to liters at STP, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (STP = 1 atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (STP = 273.15 K)
Rearranging the equation to solve for V:
V = nRT / P
V = (15.0 mol)(0.0821 L·atm/mol·K)(273.15 K) / (1 atm)
V ≈ 331.99 L
Therefore, approximately 332 liters of hydrogen gas (H2) would be produced by reacting 15.0 moles of sulfuric acid (H2SO4) with 15.0 moles of aluminum (Al) at STP.