Chemistry, asked by kavitha54, 1 year ago


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 Anonymous
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 AneesKakar
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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