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maks197457 [2]
2 years ago
11

Solid sodium hydroxide reacts with gaseous carbon dioxide to form solid sodium carbonate salt and liquid water. Formulate the ba

lanced chemical reaction When 1.85 mol NaOH and 1.00 mol carbon dioxide heat, how many moles of sodium carbonate can be theoretically produced? How many moles of the excess starting material remains?
Chemistry
1 answer:
Bond [772]2 years ago
5 0

Answer:

a) 0.925 mol Na2CO3 can be theoretically produced

b) 0.075 moles of the excess starting material remains

Explanation:

balaced chemical:

2 NaOH(s) + CO2(g) ↔ Na2CO3(s) + H2O(aq)

   1.85n          1.00n             Xn

moles theor. Na2CO3:

⇒ nNaCO3 = 1.85nNaOH * ( nNa2CO3 / 2nNaOH)

⇒ nNa2CO3 = 0.925nNa2CO3

moles of the excess:

⇒moles CO2 react = 1.85nNaOH * nCO2 / 2nNaOH = 0.925n CO2

⇒moles CO2 excess = 1.00n - 0.925n = 0.075n excess CO2

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If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o
sweet-ann [11.9K]

<u>Answer:</u>

<em>The molarity of the AgNO_3 solution is 4.02 \times 10^4 M </em>

<em></em>

<u>Explanation:</u>

The Balanced chemical equation is

1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)

Mole ratio of AgNO_3 : KCl is 1 : 1

So moles AgNO_3  = moles KCl

Moles KCl = \frac {mass}{molarmass}

= \frac {0.785 mg}{(39.1+35.5 g per mol)}

= \frac {0.000785 g}{74.6 g  per mol}

= 0. 0000105 mol KCl

= 0.0000105 mol AgNO_3

So  Molarity

= \frac {moles of solute}{(volume of solution in L)}

= \frac {0.0000105 mol}{26.2 mL}

=\frac {0.0000105 mol}{0.0262 L}

= 0.000402M or mol/L is the Answer

(Or) 4.02 \times 10^4 M is the Answer

6 0
3 years ago
How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

4 0
3 years ago
a sample of gas measures 5 liters at 1 atm. to change the volume to a 3.5 liters at constant temperature what pressure must be a
Scorpion4ik [409]

Answer:

0.7atm

Explanation:

3.5/5 because constant temperature

8 0
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1. Which of the following is an example of a physical change to matter?
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