Chemistry, 15.11.2020 23:40 haileyjones732

What mass of sodium carbonate is required for complete reaction with 8.35 g of nitric acid to produce sodium nitrate, carbon dioxide, and water?

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The mass of sodium carbonate that is required for complete reaction with 8.35 grams of nitric acid is equal to 7.02 grams.

Given the following data:

Mass of nitric acid = 8.35 grams

Scientific data:

Molar mass of HNO_3

= 63 g/mol.Molar mass of

= 106 g/mol.

To determine the mass of sodium carbonate that is required for complete reaction with 8.35 grams of nitric acid:

First of all, we would write the properly balanced chemical equation for this chemical reaction:

$Na_2CO_3 + 2HNO_3 \rightarrow 2NaNO_3 + H_2O + CO_2$

Next, we would find the number of moles of nitric acid ( HNO_3 ) required:

$Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{8.35}{63 }$

Number of moles = 0.1325 moles.

By stoichiometry:

2 moles of HNO_3 = 1 mole of Na_2CO_3

0.1325 moles of HNO_3 = X mole of Na_2CO_3

Cross-multiplying, we have:

$2X=0.1325\\\\X=\frac{0.1325}{2}$

X = 0.0663 moles

Now, we can determine the mass of sodium carbonate that is required:

$Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.0663 \times 106$

Mass of HNO_3 = 7.02 grams

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Answer from: leilalony2660

Mass of sodium carbonate = 7.022 g

Further explanation

Reaction

Na₂CO₃ + 2HNO₃ ⇒ 2NaNO₃ + H₂O + CO₂

mol of HNO₃ (MW=63,01 g/mol):

$\tt mol=\dfrac{8.35}{63.01}=0.1325$

mol Na₂CO₃

$\tt \dfrac{1}{2}\times 0.1325=0.06625$

mass Na₂CO₃(MW = 105,9888 g/mol) :

$\tt 0.06625\times 105,9888 g/mol=7.022~g$

Answer from: Quest

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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