Answer:
HCl is the limiting reactant. It will completely be consumed (1.37 moles)
Option D is correct
Explanation:
Step 1: Data given
Mass of Zinc (Zn) = 50.0 grams
Mass of Hydrogen chloride (HCl) = 50.0 grams
atomic mass Zn = 65.38 g/mol
Molar mass HCl = 36.46 g/mol
Step 2: The balanced equation
Zn + 2HCl → ZnCl2 + H2
Step 3: Calculate moles
Moles = mass / molar mass
Moles Zn = 50.0 grams / 65.38 g/mol
Moles Zn = 0.764 moles
Moles HCl = 50.0 grams / 36.46 g/mol
Moles HCl = 1.37 moles
Step 4: Calculate limiting reactant
For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2
HCl is the limiting reactant. It will completely be consumed (1.37 moles)
Zn is in excess. There will react 1.37/2 = 0.685 moles
There will remain 0.764 -0.685 = 0.079 moles
Answer:
We need 41.8 mL of NaOH
Explanation:
<u>Step 1:</u> Data given
Mass of H2X = 0.1873 grams
Molarity of NaOH solution = 0.1052 M
Molar mass of H2X = 85.00 g/mol
<u>Step 2</u>: The balanced equation
H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)
<u>Step 3:</u> Calculate moles of H2X
Moles H2X = mass H2X / Molar mass H2X
Moles H2X = 0.1873 grams / 85.00 g/mol
Moles H2X = 0.0022 moles
<u>Step 4:</u> Calculate moles of NaOH
For 1 mol H2X we need 2 moles NaOH to produce 1 mole of Na2X and 2 moles of H2O
For 0.0022 moles of H2X we need 0.0044 moles of NaOH
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<u>Step 5</u>: Calculate volume of NaOH
Volume of NaOH = moles of NaOH / molarity of NaOH
Volume of NaOH = 0.0044 moles / 0.1052 M
Volume NaOH = 0.0418 L = 41.8 mL
We need 41.8 mL of NaOH
<span>H2C2O4(aq) + 2OH- --> C2O4^2- + 2H2O(l)</span>
Answer:
The answer is 2.03
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ {H}^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7BH%7D%5E%7B%2B%7D%20%5D)
where H+ is the Hydrogen ion concentration
From the question we have
We have the final answer as
<h3>2.03 </h3>
Hope this helps you
