All categories

2 years ago

Balance the chemical equation hcl+ Naoh

Chemistry

1 answer:

sattari [20]2 years ago

6 0

Answer:

NaCL + H2O

Explanation:

You might be interested in

Why is acylation reagent important.

shepuryov [24]

Answer:

Acylation can be used to prevent rearrangement reactions that would normally occur in alkylation.

3 0

2 years ago

SOMEBODY PLEASE HELP.. I DONT UNDERSTAND

elena-s [515]

Na,C12 is the answer

4 0

2 years ago

The only positive ion found in H2SO4(aq) is the

Hitman42 [59]

The answer is ammonium iron

3 0

3 years ago

All matter can be classified by their physical properties. Which substance can be classified as being soluble in water

My name is Ann [436]

The substance that can be classified as being soluble in water would be salt. The solution is D. Salt specifically is composed of ions and the ionic compound dissociates within water.

4 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

Answer: The mass of lead iodide produced is 9.22 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago