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3 years ago

What is the concentration in molL-1 of 20 mL of Na2CO3 solution that neutralises 15 mL of 1 M HCl solution?

Chemistry

1 answer:

lutik1710 [3]3 years ago

3 0

Answer:

$0.75\ \text{M}$

Explanation:

$C_1$ = Concentration of $Na_2CO_3$

$C_2$ = Concentration of $HCl$ = 1 M

$V_1$ = Volume of $Na_2CO_3$ = 20 mL

$V_2$ = Volume of $HCl$ = 15 mL

We have the relation

$V_1C_1=V_2C_2\\\Rightarrow C_1=\dfrac{V_2C_2}{V_1}\\\Rightarrow C_1=\dfrac{15\times 1}{20}\\\Rightarrow C_1=0.75\ \text{M}$

The concentration of $Na_2CO_3$ is $0.75\ \text{M}$ .

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The metal sodium is obtained by electrolysis of common table salt (sodium chloride, nacl). how much sodium chloride needs to be

Bogdan [553]

Solve first for the number of moles of sodium given that the mass is 2500 g by dividing the given mass by the molar mass of sodium.
moles sodium = 2500 g / 23 g/mol
moles sodium = 108.7 moles
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4 years ago

6.0 mol NaOH reacts with

lina2011 [118]

Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

NaOH: 3 moles
H₃PO₄: 1 mole
H₂O: 3 moles
Na₃PO₄: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

$moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}$

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

$moles of Na_{3}P O_{4} =\frac{6 moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}$