1. Indicate the concentration of each ion present in the solution formed by mixi

Question

1. Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive).

(a) 50 mL of 0.100 M HCl and 10.0 mL of 0.560 M HCl

H+ = ______ M

Cl- = ______ M

(b) 15.0 mL of 0.292 M Na2SO4 and 20.8 mL of 0.200 M KCl

Na+ = _____ M

K+ = _____ M

SO42- = _____ M

Cl- = _____ M

(c) 3.50 g of NaCl in 51.7 mL of 0.215 M CaCl2 solution

Na+ = _____ M

Ca2+ = ______ M

Cl- = ______ M

2. Consider the balanced chemical reaction shown below.

In a certain experiment, 4.113 g of C3H6(g) reacts with 1.364 g of O2(g).

(a) Which is the limiting reactant? (Example: type C3H6 for C3H6(g))
________ is the limiting reactant.

(b) How many grams of CO2(g) form?
_______ g of CO2(g) form.

(c) How many grams of H2O(l) form?
_______ g of H2O(l) form.

(d) How many grams of the excess reactant remains after the limiting reactant is completely consumed?
______ g of excess reactant remain.

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st one for you. Please ask other as different question.

1.
a)
50 mL of 0.100 M HCl and 10.0 mL of 0.560 M HCl
number of moles of HCl from 50 mL of 0.100 M HCl = M* V = 0.1 M * 0.05 L = 0.005 mol
number of moles of HCl from 10.0 mL of 0.560 M HCl= M* V = 0.56 M * 0.01 L = 0.0056 mol

Total number of moles of HCl = 0.005 + 0.0056 mol = 0.0106 mol

Total volume = 50 mL + 10 mL = 60 mL = 0.06 L

[HCl] = number of moles / volume = 0.0106/0.06 = 0.18 M

HCl -----> H + + Cl-
[H+] = 0.18 M
[Cl-] = 0.18 M

b)
15.0 mL of 0.292 M Na2SO4 and 20.8 mL of 0.200 M KCl
number of moles of Na+ = 2*0.292 M * 0.015 L = 0.00876 mol
number of moles of SO42- = 0.292 M * 0.015 L = 0.00438 mol
number of moles of K+ =0.2 M* 0.0208 L = 0.00416 mol
number of moles of Cl- =0.2 M* 0.0208 L = 0.00416 mol

Total volume = 0.015 + 0.0208 = 0.0358 L

[Na+] = 0.00876/0.0358 = 0.245 M
[SO42-] = 0.00438 /0.0358 = 0.122 M
[K+] = 0.00416 /0.0358 = 0.12 M
[Cl-] = 0.00416 /0.0358 = 0.12 M

c)
number of moles of CaCl2 = M*V = 0.215 M * 0.0517 L = 0.011 mol
number of moles of NaCl = mass/ molar mass = 3.5/58.5 =0.06 mol

Volume = 0.0517 L

[Ca2+] = 0.011/0.0517 = 0.215 M
[Cl-] = (2*0.011 + 0.06)/0.0517 = 1.59 M
[Na+]   = 0.06/0.0517 = 1.16 M

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