Answer:
sodium hydroxide is the limiting reactant
Explanation:
The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.
2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr
Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.
For nitrogen tribromide
From the balanced reaction equation;
2 moles of nitrogen tribromide yields 1 mole of nitrogen gas
4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas
For sodium hydroxide;
3 moles of sodium hydroxide yields 1 mole of nitrogen gas
5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas
Therefore, sodium hydroxide is the limiting reactant.
Answer is: the percent purity of the sodium bicarbonate is 56.83 %.
1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.
2. m(NaHCO₃) = 3.50 g
n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).
n(NaHCO₃) = 3.50 g ÷ 84 g/mol.
n(NaHCO₃) = 0.042 mol.
3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0.042 mol.
m(CO₂) = 0.042 mol · 44 g/mol.
m(CO₂) = 1.83 g.
4. the percent purity = 1.04 g/1.83 g ·100%.
the percent purity = 56.8 %.
<span>Get a periodic table of elements. ...Find your element on the periodic table. ...Locate the element's atomic number. ...Determine the number of electrons. ...Look for the atomic mass of the element. ...<span>Subtract the atomic number from the atomic mass.</span></span>
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