Full Question;
What volume of a 0.150 M solution of KOH must be added to 450.0 mL of the acidic solution of 300ml of 0.450M HCL to completely neutralize all of the acid? Express the volume in liters to three significant figures.
Answer:
0.9l
Explanation:
First thing's first, we have to write out the balanced chemical equation.
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
Potassium hydroxide, KOH, and hydrochloric acid, HCl, react in a 1:1 mole ratio to produce aqueous potassium chloride, KCl, and water.
From the reaction;
Na = Nb
Where Na = Number of moles of acid
Na = Ca * Va = 0.450 * 0.300 = 0.135
Nb = Cb * Vb = Cb * 0.150
Na = Cb * 0.150
0.135 = Cb * 0.150
Cb = 0.135 / 0.150 = 0.9L
Answer:
B Group Salt (Ionic Bond)
Explanation:
Iron (III) Oxide
Fe2O3
Answer:
If I could have a super power it would be invisibility. Sometimes you wish you weren't there and if you were hiding from someone then they couldn't find you.
Answer:
Number of moles of hydrogen gas produced when 15.3 g of sodium reacts with water = 0.333 moles of hydrogen gas
Explanation:
The reaction between sodium metal and water is given by the chemical equation below:
2Na(s) + 2H₂O(l) ------> 2NaOH(aq) + H₂(g)
From the equation of reaction above, 2 moles of sodium reacts with 1 mole of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.
mole ratio of sodium and hydrogen gas is 2:1
molar mass of sodium =23 g/mol:
number of moles of sodium present in 15.3 g = mass/molar mass
number of moles of sodium present in 15.3 g = 15.3 g/ 23 g/mol = 0.665 moles
number of moles of hydrogen gas produced = 0.665 * 1/2 = 0.333 moles
Therefore, number of moles of hydrogen gas produced when 15.3 g of sodium reacts with water is 0.333 moles