Left Panel
Short answer A
Solution
Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.
Equation
Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.
Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2
Step One
Balance the Equation
2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2
Step Two
Find the molar mass of CaCl2
Ca = 40
2Cl = 71
Molar Mass = 40 + 71 = 111 grams/mole
Step Three
Find the number of moles of CaCl2
Given mass = 379.4
Molar Mass = 111
moles = given Mass / molar Mass
moles of CaCl2 = 379.4/111 = 3.418 moles
Step Four
Find the number of moles of Ca3(PO4)2 needed.
This requires that you use the balance numbers from the balanced equation.
For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2
n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles
Step Five
Find the molar mass of Ca3(PO4)2
From the periodic table,
3Ca = 3 * 40 = 120
2 P = 2 * 31 = 62
8 O = 8 * 16 =128
Molar Mass = 120 + 62 + 128= 310 grams per mole.
Step Six
1 mole of Ca3(PO4)2 has a molar mass of 310 gram
1.13933 moles of Ca3(PO4)2 = x
x = 1.13933 moles * 310 grams /mole
x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.
Middle Panel
Short Answer C
Equation
2HCl + Mg ===> H2 + MgCl2
The object of the first part of the game is to find the number of moles of H2.
Step One
Find the moles of HCl
1 mole HCl = 35.5 + 1 = 36.5
n = given mass divided by molar mass
n = 49 grams / 36.5 = 1.34 moles.
The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.
So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.
Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.
Find V using PV = nRT
R is going to be in kPa so the value of R = 8.314
V = ???
n = 0.671 moles
T = 25 + 273 = 298oK
P = 101.3 kPa
101.3 * V= 0.671*8.314 * 298
V = 0.671 * 8.314 * 298 / 101.3
V = 16.4
The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.
Right Panel
I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.
LiOH + HBr ===> LiBr + H2O and the equation is balanced.
You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles
n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles
n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles
HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.
n = given mass / molar mass multiply both sides by the molar mass
n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of
The answer is C