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2 years ago

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A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo

lume of 1.67 L at 1.24 atm and 26 degree C. From these data, calculate the percent composition by mass of Na2CO3 in the mixture

1 answer:

Svetlanka [38]2 years ago

3 0

Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %

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The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.

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At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol

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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a

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Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

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The general term for a large molecule made up of many similar subunits is

Setler79 [48]

A) Polymer is the general name of large units made of many smaller units (these would be called monomers). An example is starch, this is a carbohydrate polymer that is made up of smaller units (monomers) called glucose.

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A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in

Vladimir79 [104]

Answer:

A

Explanation:

Considering question A

Mass of original sample is $m_o = 0.945 \ g$

Mass of NH4Cl is $m_n = 0.116 \ g$

Percent of NH4Cl is $k = 12.275 \%$

B

Mass of NaCl $m_k = 0.359 \ g$

C

Mass of SiO2 $m_e = 0.46$

D

Mass of original sample $m_o = 0.945 \ g$

Differences in these weights (g) (use the absolute value of the difference)

recovery of matter $G = 0.01 \ g$

The correct option is C

From the question we are told that

The mass of evaporating dish on #1 is $m_1 = 38.646 \ g$

The mass of evaporating dish and original sample $m_2 = 39 591 \ g$

The mass of evaporating dish after subliming $NH_4Cl$ is $m_3 = 39.4750 \ g$

Generally the mass of the original sample is mathematically represented as

$m_o = m_2 - m_1$

=> $m_o = 39 591 - 38.646$

=> $m_o = 0.945 \ g$

Generally the mass of $NH_4Cl$ is mathematically represented as

$m_n = m_2 - m_3$

=> $m_n = 39 591 - 39.4750$

=> $m_n = 0.116 \ g$

The Percent $NaH_4 Cl (g)$

$k = \frac{ m_n}{m_o} *100$

=> $k = \frac{0.116 }{0.945} *100$

=> $k = 12.275 \%$

Considering question B

The mass of evaporating dish #2 is $m_g = 38700\ g$

The mass of watch glass is $m_a = 28 299 \ g$

The mass of evaporating dish #2, watch glass and NaCl $m_b = 67,355 \ g$

Generally the mass of NaCl is

$m_k = m_b -[m_g + m_a]$

=> $m_k = 67,355 -[38700 + 28 299]$

=> $m_k = 0.359 \ g$

Considering question C

The mass of evaporating dish is $m_p= 38.645$

The mass of evaporating dish and SiO2 $m_s = 39.105 \ g$

Generally the mass of SiO2 is mathematically represented as

$m_e = 39.105 - 38.645$

=> $m_e = 0.46$

Considering D

The mass of the original sample is $m_o = 0.945 \ g$

Generally the experimental mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

$M =0.116 + 0.46 + 0.359$

$M = 0.935 \ g$

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

$G =0.945- 0.935$

=> $G = 0.01 \ g$

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

6 0

2 years ago