The balanced equation for the reaction between Na and H₂SO₄ is as follows;
2Na + H₂SO₄ ---> Na₂SO₄ + H₂
Stoichiometry of Na to H₂SO₄ is 2:1
the molar concentration of H₂SO₄ is 6.0 M
This means that in 1 L of H₂SO₄ - there are 6.0 mol of H₂SO₄
Therefore in 750 mL -
Number of H₂SO₄ moles reacted - 4.5 mol
According to stoichiometry,
1 mol of H₂SO₄ reacts with 2 mol of Na
Therefore 4.5 mol reacts with 2x4.5= 9 mol of Na
Mass of Na reacted - 9 mol x 23 g/mol = 207 g of Na is required
The number of moles of AlBr3 = Molarity × Volume
= 0.150 × 0.3
= 0.045 moles
1 mole of AlBr3 gives 3 moles of Br- ions
That is
AlBr3 = Al3+ + 3Br-
Therefore, 0.045 moles of AlBr3 will yield 3 × 0.045 = 0.135 moles
Thus; they are 0.135 moles of bromide ions
Answer:
0.109 M
Explanation:
change the 974 mL to L (0.974L)
17g x molar mass (159.6g/mol)/0.974=0.109 M
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