A gaseous mixture containing 5 moles of H2 and 7 moles of Br2 reacts to form HBr. Write a balanced chemical equation. Which reactant is limiting? What is the theoretical yield in moles? What is the theoretical yield for this reaction in grams? How many moles of excess reactant are left over? How many grams of excess reactant are leftover?
To write a balanced chemical equation for the reaction between H2 and Br2 to form HBr, you need to consider the law of conservation of mass. The number of atoms of each element must be equal on both sides of the equation.
The balanced chemical equation is:
H2 + Br2 -> 2HBr
From the equation, you can see that 1 mole of H2 reacts with 1 mole of Br2 to produce 2 moles of HBr.
To determine the limiting reactant, you need to compare the mole ratios of the reactants with their actual amounts.
Given the initial amounts, there are 5 moles of H2 and 7 moles of Br2.
To find the limiting reactant, you need to convert the moles of each reactant to the mole ratio with respect to the desired product (HBr) using the coefficients in the balanced equation.
H2:Br2 = 5 moles : 1 mole = 5 : 1
HBr:Br2 = 2 moles : 1 mole = 2 : 1
Comparing these ratios, you can see that the mole ratio of H2:Br2 is higher than the mole ratio of HBr:Br2. Therefore, Br2 is the limiting reactant.
The theoretical yield in moles is the maximum amount of HBr that can be produced from the limiting reactant. Since 1 mole of Br2 reacts to form 2 moles of HBr, and Br2 is the limiting reactant with a quantity of 7 moles, the theoretical yield of HBr is 7 moles * 2 moles HBr / 1 mole Br2 = 14 moles of HBr.
To calculate the theoretical yield in grams, you need to know the molar mass of HBr. The molar mass of HBr is approximately 80.91 g/mol. Therefore, the theoretical yield in grams is 14 moles of HBr * 80.91 g/mol = 1132.74 grams of HBr.
To determine the moles of excess reactant left over, you can use the mole ratio between the limiting reactant (Br2) and the excess reactant (H2).
From the balanced equation, you can see that 1 mole of H2 reacts with 1 mole of Br2. Since Br2 is the limiting reactant with a quantity of 7 moles, there must be an equal number of moles of H2 leftover. Therefore, the moles of excess reactant (H2) left over is also 7 moles.
To calculate the grams of excess reactant left over, you need to know the molar mass of H2. The molar mass of H2 is approximately 2.02 g/mol. Therefore, the grams of excess reactant left over is 7 moles of H2 * 2.02 g/mol = 14.14 grams of H2.