1. What volume of H 2 gas (in liters), measured at 27°C and 670. torr, can be ob
ID: 692352 • Letter: 1
Question
1. What volume of H2 gas (in liters), measured at 27°C and 670. torr, can be obtained by reacting 5.00 g of zinc metal with 189 mL of 0.224 M HCl? The equation is
2. Determine how many grams of Al(OH)3 will be required to neutralize 270 mL of 0.245 M HCl according to the reaction:
3HCl(aq) + Al(OH)3(s) AlCl3(aq) + 3H2O(l)
3. A paint requires 25.0 g of iron(III) oxide to give it the right yellow tint. How many grams of a 30.0% by mass solution of iron(III) oxide should be used?
_____ g solution
4. In 60. g of a 17.2 % by mass solution of BaCl2:
B). how many grams of solvent are present? _______ g solvent
5. Some people have begun to formulate their own window-cleaning solutions using commonly available household chemicals. One formulation calls for mixing vinegar (dilute acetic acid, HC2H3O2) with water and a small amount of detergent. A commonly used recipe contains 10.8 g of acetic acid mixed with 574 g of water and 3.3 g of detergent. This recipe makes a total of 588 mL of solution.
A). What is the mass percent of acetic acid in the window cleaner? _____%
B). What is the molarity of acetic acid in the window cleaner? ______M C2H4O2
Please, I need help with these questions. Thank You
Explanation / Answer
Answer-(1)
(1) Write the balanced equation: Zn(s)+2HCl(aq)--->ZnCl2(aq)+H2(g)
(2) calc. moles or reactants:
moles Zn(s) = 5.00 g x 1 mol/65.4 g = 0.076 moles Zn
moles HCl =0.189Lx 0.224 mol/L =0.042 moles of HCL
3) Determine limiting reactant: not enough HCl to react with 0.076 moles Zn.
Need 2 x 0.076 moles.
Thus, HCl is limiting.
4) Determine moles of H2(g) produced: 2 moles HCl = 1 mole H2(g) ...
so 0.042 moles HCl =( ½ )x (0.042)= 0.21 moles H2 gas.
5) Convert moles of gas to volume of gas: PV = nRT and solve for V = nRT/P
V = (0.021) (0.0821) (300 deg K) /670/760
(I changed torr to atm by dividing by 760 and deg C to K by adding 273)
V = 0.586 liters
Answer-(2)
You look at the balanced equation, you will see that it takes one mole of Al(OH)3 to balance three moles of HCl.
we must find how many moles of HCl. Once we do that, we'll know we need 1/3rd as many moles of Al(OH)3 to completely react with the acid.
0.270 liters (of HCl) x 0.245 moles/liter = 0.0661 moles
so we need 1/3 x 0.0661
0.0220 moles of Al(OH)3 to do the job.
Finally, how many grams is that?
0,0220 moles x 78g/mole = 1.71 grams
Answer (4)
solute is BaCl2
solvent is x
17.2 % x 60 gives you how many grams of BaCl2 are in the solution which is 10.32 grams
82.8% remains which would be the percent of solvent so do the same as before.
82.8% x 60g gives 49.68 g of solvent
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