<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
Answer:
Explanation:
Chemical Equations are representations of chemical reactions in terms of the symbols and formulae of the elements and compounds involved. A chemical equation usually have the reactant at the left hand side while the product is on the right hand side.
A chemical Equation is of little or no value if is not in balanced equation. When an equation is balanced , the total number of atoms of any element on the left-hand side of it must be equal to the total number of atoms of that element on the right hand side.
in the given question; we are given a word problem of chemical symbol to compute and also to balance the chemical equation.
From below; the chemical equation can be written as:
From the above equation we will notice that it is not truly balanced ; so th balanced equation can be written as:
Answer:
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N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹
n(Mg)=N/Nₐ
m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ
m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
