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 %.
Explanation:
it should be option number (D) Iron (III) oxide
Molar mass of CuSO4 * 5 H2O
= 63.546 + 32 + 16*4 + 5*18
= 249.546 g/mol
Mass of water in that formula: 5 * 18 = 90 g/mol
Percent by mass of water = 90 / 249.546 = 36%
<span>So, 36% of your 8.22 g is water. 0.36 * 8.22= 2.95 g of water
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We know that acids have a pH of under 7.
We also need to:
Set up an ICE table for the chemical reaction. Solve for the concentration of H3O+ using the equation for pH Use the concentration of H3O+ to solve for the concentrations of the other products and reactants.
