Answer:
Mass = 245.72 g
Explanation:
Given data:
Mass of SrCl₂ react = ?
Mass of H₂SO₄ = 300.0 g
Solution:
SrCl₂ + 2H₂SO₄ → 2HCl + Sr(HSO₄)₂
Number of moles of H₂SO₄:
Number of moles = mass/molar mass
Number of moles = 300.0 g/ 98.079 g/mol
Number of moles = 3.1 mol
Now we will compare the moles of SrCl₂ and H₂SO₄.
H₂SO₄ : SrCl₂
2 : 1
3.1 : 1/2×3.1 = 1.55 mol
Mass of SrCl₂:
Mass = number of moles × molar mass
Mass = 1.55 mol × 158.53 g/mol
Mass = 245.72 g
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
Answer:
Explanation:
This is a bond with reacts with metal's