1.00 g of alkali metal carbonate, $M_{2} C O_{3}$ $M_2CO_3$ , is dissolved in water and made up to $250 c m^{3}$ $250 cm^3$ in a volumetric flask. $25.00 c m^{3}$ $25.00 cm^3$ portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is $12.80 c m^{3}$ $12.80cm^3$ . What element is M?

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1.00 g of alkali metal carbonate, $M_{2} C O_{3}$ $M_2CO_3$ , is dissolved in water and made up to $250 c m^{3}$ $250 cm^3$ in a volumetric flask. $25.00 c m^{3}$ $25.00 cm^3$ portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is $12.80 c m^{3}$ $12.80cm^3$ . What element is M?

1.00 g of alkali metal carbonate, $M_{2} C O_{3}$ $M_2CO_3$ , is dissolved in water and made up to $250 c m^{3}$ $250 cm^3$ in a volumetric flask.
$25.00 c m^{3}$ $25.00 cm^3$ portions of this solution are titrated against 0.113M hydrochloric acid. The average titre is $12.80 c m^{3}$ $12.80cm^3$
Calculate:
a. The moles of HCl in the titre
b. The moles of $M_{2} C O_{3}$ $M_2CO_3$ in 25 cm3 of solution
c. The moles of $M_{2} C O_{3}$ $M_2CO_3$ in 250 cm3 of solution
d. The relative formula mass of $M_{2} C O_{3}$ $M_2CO_3$
e. The identity of M

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Answer 1 · 2017-11-14T20:34:51

One necessary initial step is to write the stoichiometric equation....

$M_{2} C O_{3} (s) + 2 H C l (a q) \to 2 M C l (a q) + H_{2} O (l) + C O_{2} (g) ↑ ⏐$ $M_2CO_3(s) + 2HCl(aq) rarr2MCl(aq) + H_2O(l) + CO_2(g)uarr$

Explanation:

$a. Moles of HCl in titre:$ $"a. Moles of HCl in titre:"$

$12.8 \cdot c m^{3} \times 10^{- 3} \cdot L \cdot c m^{- 3} \times 0.113 \cdot m o l \cdot L^{- 1} = 1.45 \times 10^{- 3} \cdot m o l$ $12.8*cm^3xx10^-3*L*cm^-3xx0.113*mol*L^-1=1.45xx10^-3*mol$

$b. Moles of carbonate in 25 mL aliquot:$ $"b. Moles of carbonate in 25 mL aliquot:"$

$\frac{12.8 \cdot c m^{3} \times 10^{- 3} \cdot L \cdot c m^{- 3} \times 0.113 \cdot m o l \cdot L^{- 1}}{2} = 7.23 \times 10^{- 4} \cdot m o l$ $(12.8*cm^3xx10^-3*L*cm^-3xx0.113*mol*L^-1)/2=7.23xx10^-4*mol$

$c. Moles of carbonate in 250 mL:$ $"c. Moles of carbonate in 250 mL:"$

$7.23 \times 10^{- 4} \cdot m o l \times 10 = 7.23 \times 10^{- 3} \cdot m o l$ $7.23xx10^-4*molxx10=7.23xx10^-3*mol$

$d. Formula mass of carbonate:$ $"d. Formula mass of carbonate:"$

$= \frac{1.00 \cdot g}{7.23 \times 10^{- 3} \cdot m o l} = 138.3 \cdot g \cdot m o l^{- 1}$ $=(1.00*g)/(7.23xx10^-3*mol)=138.3*g*mol^-1$

$e. Identity of metal and metal carbonate:$ $"e. Identity of metal and metal carbonate:"$

We know that the formula mass of $M_{2} C O_{3} \equiv 138.3 \cdot g \cdot m o l^{- 1}$ $M_2CO_3-=138.3*g*mol^-1$ .

Since the mass of $C O_{3}^{2 -}$ $CO_3^(2-)$ is $60 \cdot g \cdot m o l^{- 1}$ $60*g*mol^-1$ , the mass of the metal is.... $\frac{138.3 - 60.0 \cdot g \cdot m o l^{- 1}}{2} = 39.1 \cdot g \cdot m o l^{- 1}$ $(138.3-60.0*g*mol^-1)/2=39.1*g*mol^-1$ .

And a quick glance at the Periodic Table confirms that we gots potassium metal, and the carbonate is $K_{2} C O_{3}$ $K_2CO_3$ .

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