When 3.00g of an anhydrous nitrate of a Group 2 metal is decomposed, 1.53g of gas is
produced.
What is the nitrate compound?
Answers
Answer:
Decomposition of anhydrous Group II nitrates-
2X(NO
3
)
2
⟶2XO+4NO
2
+O
2
Here, X represents the Group II metal
There are two types of gases produced. Hence it is not possible to find the number of moles of the gases produced with the information given.
Therefore,
Amount of solid metal oxide =3−1.53=1.47g
From the above equation,
2 moles of the metal nitrate will produce 2 moles of the metal oxide. Hence it is a 1:1 mole ratio.
To solve this question, the molecular mass of each metal nitrate and metal oxide must be calculated. The answer would be the compound which has equal number of moles of metal nitrate and metal oxide for the masses given.
Therefore,
Number of moles of strontium nitrate =
211.6
3
=0.014 mol
Number of moles of strontium oxide =
103.6
1.47
=0.014 mol
Answer:
3.00g of an anhydrous nitrate when gorup of metal 2×1.53of gas
3.00 anhydrous nitrate -3.53
=0.53 of nitrate compuond