47.) What is the percentage purity of magnesium
carbonate sample whose 8.4 g sample on
decomposition gives 2.2 g Co,?
(1) 25%
MgCO3
(2) 100%
for 50%
(4) 75%
Answers
Answered by
0
Answer:
MgCO₃ ⇒ MgO + CO₂
84g 40g
y g 8 g (y is variable used)
y=(84 x 8)/40 = 16.8 g
Percentage purity of MgCO₃ = (16.8 x 100)/20=84%
Answered by
4
Answer:
50% pure.
Explanation:
When decomposition of magnesium carbonate occurs we get the reaction as given below:-
MgCO3 -> MgO + CO2.
Hence, form the above we get that. equal number of moles of CO2 are formed as that of MgCO3.
So, the effective mass of MgCO3 is given by formula -
n(MgCO3) = n(CO2).
W / 84 = 2.2 / 44.
W = 84 / 20.
W = 4.2 g.
Again, the percentage purity of MgCO3 can be solved by:-
Percentage purity = (effective mass / given mass) × 100.
Percentage purity = 4.2 / 8.4 × 100.
Percentage purity = 50 %.
So, we can conclude that the percentage purity of the MgCO3 is 50%.
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