Mass of sodium carbonate = 7.022 g
Reaction
Na₂CO₃ + 2HNO₃ ⇒ 2NaNO₃ + H₂O + CO₂
mol of HNO₃ (MW=63,01 g/mol):
[tex]\tt mol=\dfrac{8.35}{63.01}=0.1325[/tex]
mol Na₂CO₃
[tex]\tt \dfrac{1}{2}\times 0.1325=0.06625[/tex]
mass Na₂CO₃(MW = 105,9888 g/mol) :
[tex]\tt 0.06625\times 105,9888 g/mol=7.022~g[/tex]
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:
Scientific data:
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:
[tex]Na_2CO_3 + 2HNO_3 \rightarrow 2NaNO_3 + H_2O + CO_2[/tex]
Next, we would find the number of moles of nitric acid ([tex]HNO_3[/tex]) required:
[tex]Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{8.35}{63 }[/tex]
Number of moles = 0.1325 moles.
By stoichiometry:
2 moles of [tex]HNO_3[/tex] = 1 mole of [tex]Na_2CO_3[/tex]
0.1325 moles of [tex]HNO_3[/tex] = X mole of [tex]Na_2CO_3[/tex]
Cross-multiplying, we have:
[tex]2X=0.1325\\\\X=\frac{0.1325}{2}[/tex]
X = 0.0663 moles
Now, we can determine the mass of sodium carbonate that is required:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.0663 \times 106[/tex]
Mass of [tex]HNO_3[/tex] = 7.02 grams
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