Element M(a solid) reacts with an aqueous solution of compound B to form an aqueous solution of compound C(a salt) and a gas composed of element D. When 1.000g of element M is reacted with excess oxgen 1.658g of the compound MO are formed. Compound B is binary, monoprotic acid (hydrogen halide) HX. A student dissolves 1.000g of HX in enough water to make 50.00ml of solution and titrates it with 0.5000 M KOH. This requires 54.79ml of base to reach end point. Compound C is binary salt that contains elements M and X. A 3.000g sample of element D at STP occupies a volume of 33.3L. A student places 1.05g of M in 50.00ml of 1.00M HX. Assume the reaction goes to completion with 100% yield. (1.) Identify M, B, D, and X. Write a balanced equation for the reaction between M and HX.(2.) What mass of compound C will be formed, and what molar concentration will it have in the final mixature. ( Assume that the volume of the final mixture is 50.00 ml). (3.)Gas D is collected by water displacement at 23 degree C and an atmospheric pressure of 755torr. What volume should be collected? (4.) How many moles of which reactant are left over when the reaction is completed?
To solve this problem, we will go step by step.
(1.)
The information given states that element M reacts with compound B (HX) to form compound C, which is a binary salt that contains elements M and X. The gas produced in the reaction is element D.
From the information provided, we know that compound B is a binary, monoprotic acid (hydrogen halide). Therefore, B must be either HCl, HBr, or HI.
To find out the identity of M, we need to use the stoichiometry of the reaction. The mass of compound MO formed is given as 1.658g. The molar mass of MO can be calculated using the molar mass of M and the mass of oxygen (16 g/mol).
Let's calculate the molar mass of M:
Molar mass of M = Mass of MO - Mass of Oxygen
= 1.658 g - 16 g/mol
= 1.658 g - 16 g/mol
Therefore, the molar mass of M is 47.004 g/mol.
Looking at the periodic table, we find that the closest molar mass to 47.004 g/mol belongs to the element Ag (silver). Thus, M = Ag.
Now, let's write the balanced equation for the reaction between M (Ag) and HX:
Ag + HX → AgX + H₂↑
(2.)
To determine the mass of compound C formed, we need to use stoichiometry once again. We are given that 1.05g of M reacts with 1.00M HX in a 50.00ml solution.
First, let's convert the volume of the solution to moles of HX.
Moles of HX = Volume of HX solution (in liters) × Molarity of HX
= 50.00 ml ÷ 1000 (to convert to liters) × 1.00 mol/L
= 0.0500 mol
Using the balanced equation, we can see that the ratio of Moles of M to Moles of HX to Moles of C is 1:1:1.
So, the mass of compound C is equal to the molar mass of C (which can be calculated from the molar mass of M and X) multiplied by the number of moles of C.
Let's calculate the molar mass of C:
Molar mass of C = Molar mass of M + Molar mass of X
= 107.868 g/mol + Molar mass of X (unknown)
To find the molar mass of X, we need more information. However, we can still calculate the mass of compound C based on the assumption that Molar mass of X is 35.453 g/mol (the molar mass of Cl, the lightest halogen).
Mass of C = Molar mass of C × Moles of C
= (107.868 g/mol + 35.453 g/mol) × 0.0500 mol
= 7.184 g
The molar concentration of C in the final mixture can be calculated by dividing the moles of C by the final volume of the mixture. We are given that the volume of the final mixture is 50.00 ml.
Molar concentration of C = Moles of C / Volume of final mixture
= 0.0500 mol / 0.0500 L
= 1.00 mol/L
Therefore, compound C will have a mass of 7.184 g and a molar concentration of 1.00 mol/L in the final mixture.
(3.)
To determine the volume of gas D collected, we need to use the ideal gas law.
Given:
Mass of D = 3.000 g
Temperature = 23°C = 23 + 273.15 = 296.15 K
Pressure = 755 torr = 755/760 atm
Using the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
We know the volume and temperature, and we need to find the number of moles (n) to solve for the volume (V) of gas D.
First, let's calculate the number of moles using the equation mol = mass / molar mass:
Molar mass of D = Mass of D / Moles of D
= 3.000 g / 47.004 g/mol (molar mass of Ag)
= 0.0638 mol
Now, let's use the ideal gas law to calculate the volume of gas D:
PV = nRT
(V) × (755/760 atm) = (0.0638 mol) × (0.0821 L·atm/(mol·K)) × (296.15 K)
(V) = (0.0638 mol) × (0.0821 L·atm/(mol·K)) × (296.15 K) / (755/760 atm)
(V) = 6.64 L
Therefore, the volume of gas D collected should be 6.64 L.
(4.)
To determine the excess/reactant remaining after the completion of the reaction, we need to calculate the moles of each reactant used and compare it with the stoichiometric ratio in the balanced equation.
From earlier calculations, we know that for the reaction between Ag (M) and HX:
1 mole of Ag requires 1 mole of HX
Given:
Mass of M = 1.05 g
Density of HX = 1.00 g/ml
The first step is to convert the mass of M to moles:
Moles of M = Mass of M / Molar mass of M
= 1.05 g / 107.868 g/mol (molar mass of Ag)
= 0.0097494 mol
To calculate the moles of HX used, we need to determine the volume of HX solution used in the titration. We are given that 0.5000 M KOH requires 54.79 ml of base to reach the end point.
Using the balanced equation for the reaction between KOH and HX, which is a neutralization reaction:
KOH + HX → KX + H₂O
The stoichiometric ratio is 1:1 between KOH and HX. Therefore, the moles of HX used equal the moles of KOH titrated.
Moles of HX used = Molarity of KOH × Volume of KOH solution used (in liters)
= (0.5000 mol/L) × (54.79 ml / 1000 ml/L)
= 0.027395 mol
Comparing the moles of M (0.0097494 mol) and HX (0.027395 mol), we can determine the limiting and excess reactants.
Since M:H₂X ratio is 1:1 in the balanced equation, we can see that M (Ag) is the limiting reactant. This means that all of the M will react with 0.0097494 mol of HX.
Therefore, the number of moles of HX remaining after the reaction is completed is:
Moles of HX remaining = Moles of HX used - Moles of HX reacting
= 0.027395 mol - 0.0097494 mol
= 0.0176456 mol
Based on the stoichiometry, we can calculate the mass of excess HX remaining by multiplying the moles of excess HX by its molar mass:
Mass of excess HX remaining = Moles of excess HX × Molar mass of HX
= 0.0176456 mol × (Mass of HX remaining / Moles of HX remaining)
= 0.0176456 mol × (Mass of HX remaining / 0.0176456 mol )
= Mass of HX remaining
Therefore, the mass of the excess HX remaining can be directly calculated as the molar mass of HX remaining.
I hope this explanation helps you understand how to solve the given problem step by step. Let me know if you have any further questions.