The interaction of metal M weighing 14 g with chlorine formed 40.6 g of the chloride of this metal.
The interaction of metal M weighing 14 g with chlorine formed 40.6 g of the chloride of this metal. Determine which metal you are talking about. Use the atomic masses of common isotopes for calculations
Decision:
Making calculations using available data
1) Write the reaction equation:
Me + Cl2 => MeCl (z);
2) To find the amount of substance Me and MeCl (z), it is necessary to know their molar masses. Therefore, we take the atomic mass of Me as (x), and the atomic mass of Cl is 35.5 (according to the table of D.I.Mendeleev). Then:
n (Me) = m (Me) / Mr (Me) = 14 / x mol;
n (MeCl (z)) = m (MeCl (z)) / Mr (MeCl (z)) = 40.6 / (x + z * 35.5);
Determination of the unknown Me by the assumed valence
3) Suppose that Me has valence I, therefore:
2Me + Cl2 => 2MeCl;
4) Make an equation (taking into account the coefficients in front of the substances in the reaction equation):
n (Me) = n (MeCl);
14 / x = 40.6 / (x + 1 * 35.5);
x = 19;
Ar = x = 19 – fluorine (F) – is not a metal;
5) Suppose that Me has valency II, then:
Me + Cl2 => MeCl2;
6) Make an equation (taking into account the coefficients in front of the substances in the reaction equation):
n (Me) = n (MeCl2);
14 / x = 40.6 / (x + 2 * 35.5);
x = 37;
Ar = x = 37 – there is no element with such an atomic mass;
7) Suppose that Me has valence III, then:
2Me + 3Cl2 => 2MeCl3;
4) Make an equation (taking into account the coefficients in front of the substances in the reaction equation):
n (Me) = n (MeCl3);
14 / x = 40.6 / (x + 3 * 35.5);
x = 56;
Ar = x = 56 – iron (Fe) – is a metal.
Answer: The unknown metal is iron (Fe).
