Can you please show me step by step on how to do these problems Directions for t
ID: 861460 • Letter: C
Question
Can you please show me step by step on how to do these problems
Directions for the Equation Set: FOR QUESTIONS 1-4: (1) Complete the word equation
and (2) Write and balance the symbolic equation, FOR QUESTIONS 5-7 Solve the
stoichiometry completely. Due Monday, December 15 in class.
1. sodium reacts with water
_____________________________________________________________________________
2. lithium hydroxide reacts with nitric acid
_____________________________________________________________________________
3. chlorine gas reacts with sodium bromide
_____________________________________________________________________________
4. copper (II) sulfate reacts with magnesium
_____________________________________________________________________________
5. Given the following equation:
BaCl2(aq) + Na2SO4(aq)
Explanation / Answer
1.
Sodium reacts with water to give Sodium hydroxide and hydrogen gas.
2Na(s) + 2H2O(l) -------------> 2NaOH(aq) + H2(g)
-------------
2.
Lithium hydroxide reacts with nitric acid to give Lithium nitrate and water
LiOH(aq) + HNO3(aq) ----------> LiNO3(aq) +H2O (l)
------------------------------
3.
Chlorine gas reacts with Sodium bromide to give Sodium Chloride and Bromine gas.
Cl2(g) + 2NaBr(s) --------------> 2NaCl(s) + Br2(g)
------------------------
4.
Copper(II)sulfate reacts with magnesium to give Magnesium sulfate and copper metal.
CuSO4(aq) + Mg(s) -----------> MgSO4(aq) + Cu(s)
-----------------------
5.
The given equation is
BaCl2(aq) + Na2SO4(aq) -----------------> BaSO4(s) + NaCl(aq)
The balanced equation is
BaCl2(aq) + Na2SO4(aq) -----------------> BaSO4(s) + 2NaCl(aq)
Given,
mass of sodium sulfate = 12.0 g
Moles of sodium sulfate = 12.0g / 142 = 0.085 mol
Mass of barium chloride = 14.10 g
Moles of Barium chloride = 14.10 / 208 = 0.068 mol.
Determination of limiting reagent:
In order to determine the limiting reagent, we need to divide the moles of each compound with their respective stoichiometric coefficient. And the one with lower number of moles is the limiting reagent.
For, sodium sulfate = 0.085 / 1 = 0.085
Barium chloride = 0.068 / 1 = 0.068
So, of the two, barium chloride has lower number of moles. Hence, Barium chloride is the limiting reagent.
A.
So, we use moles of barium chloride to calculate the precipitate formed.
From the given reaction, it is clear that 1 mol of BaCl2 gives 1 mol of BaSO4.
Therefore, 0.068 mol of BaCl2 gives 0.068 mol of BaSO4.
Moles of BaSO4 = 0.068
Mass of BaSO4 = Moles x Molar mass
=> 0.068 x 233 = 15.84 g.
Mass of precipitate = 15.84 g
B.
As, the limiting reagent is Barium chloride, here too we use the moles of Barium chloride.
From the given equation, it is clear that, 1 mol of BaCl2 gives 2 mol of Sodium chloride.
So, 0.068 mol of BaCl2 gives 2 x 0.068 mol NaCl.
Moles of NaCl = 0.136 mol.
Mass of NaCl = Moles x Molar mass
=> 0.136 x 58.44 = 7.95 g.
Mass of NaCl formed = 7.95 g
-------------------------------------
6.
Given, mass of water = 125 g (since, density of water = 1g/ml)
Here, there are five stages of heat energy release.
Q1 = energy released when steam at 183 oC is converted to steam at 100 oC
Q1 = m x c x delta T
=125 x 2.09 x 83 (C for steam = 2.09 J/g oC)
= 21,683.75 J
Q2 = energy released when steam at 100 degree C is converted to water at 100 oC.
Q2 = m x delta Hvap
=125 x 2257 (Delta Hvap = 2257 J/g)
= 282,125 J
Q3 = energy released when water at 100 oC is converted to water at 0 oC
Q3 = m x c x delta T
= 125 x 4.184 x 100 (C for water = 4.184 J/g oC)
= 52,300 J
Q4 = energy released when water at 0 oC is converted to ice at 0 oC
Q4 = m x Delta Hfus
= 125 x 334 (delta Hfus = 334 J/g)
= 41,750 J
Q5 = energy released when ice at 0 oC is converted to ice at -10 oC
Q5 = m x c x delta T
= 125 x 2.09 x 10 ( C for ice is 2.09 J/g)
= 2612.5 J
Total energy = Q1 + Q2 + Q3 + Q4 + Q5
= 21,683.75 J + 282,125 J + 52,300 J + 41,750 J + 2612.5 J
= 400,471.25 J
= 400.5 kJ