A $100 \cdot m L$ $100mL$ volume of $2.0 \cdot m o l \cdot L^{- 1}$ $2.0mol*L^-1$ hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

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A $100 \cdot m L$ $100mL$ volume of $2.0 \cdot m o l \cdot L^{- 1}$ $2.0mol*L^-1$ hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

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Answer 1 · 2017-02-16T16:09:24

The equivalent weight of the metal is $100.0 \cdot g \cdot m o l^{- 1}$ $100.0*g*mol^-1$

Explanation:

First we need a chemical equation:

$M C O_{3} (s) + 2 H C l (a q) \to M C l_{2} (a q) + H_{2} O (l) + C O_{2} (g) ↑ ⏐$ $MCO_3(s) + 2HCl(aq) rarr MCl_2(aq) + H_2O(l) + CO_2(g)uarr$

$Moles of HCl = 0.100 \cdot L \times 2 \cdot m o l \cdot L^{- 1} = 0.200 \cdot m o l$ $"Moles of HCl"=0.100*Lxx2*mol*L^-1=0.200*mol$

Given the equation, the $metal carbonate$ $"metal carbonate"$ represents $0.100 \cdot m o l$ $0.100*mol$ , and thus, given the equation, the equivalent weight of the $metal carbonate$ $"metal carbonate"$ is $100.0 \cdot g \cdot m o l^{- 1}$ $100.0*g*mol^-1$ . This equivalent weight is CONSISTENT with a formula of $C a C O_{3}$ $CaCO_3$ .

Answer 2 · 2017-10-21T01:45:45

Equivalent weight of metal is 20 g/eq

Explanation:

For HCl
Molarity = Normality

Gram equivalents of HCl = Normality × Volume in litres
$= 2 N \times 100 \times 10^{- 3} L$ $= 2 N × 100 × 10^-3 L$
$= 0.2$ $= 0.2$ eq

For complete neutralisation
Gram equivalents of HCl = Gram equivalents of metal carbonate

Equivalent weight of metal carbonate $= \frac{Weight of metal carbonate}{Gram equivalents of metal carbonate}$ $= text(Weight of metal carbonate) / text(Gram equivalents of metal carbonate)$

$= \frac{10 g}{0.2 eq} = 50$ $= (10 g) / (0.2 text(eq)) = 50$ g/eq

Equivalent weight of carbonate $(C O_{3}^{2 -}) = \frac{12 + (3 \times 16)}{2} = 30$ $(CO_3^(2-)) = (12 + (3 × 16))/2 = 30$ g/eq

Equivalent weight of metal carbonate = Equivalent weight of metal + Equivalent weight of carbonate

50 g/eq = x + 30 g/eq
x = 20 g/eq

∴ Equivalent weight of metal is 20 g/eq

Answer 3 · 2017-10-21T07:48:55

The number of moles of $H C L$ $HCL$ reacted with Carbonate salt is

$= molarity of acid \times volume of acid in L$ $="molarity of acid" xx " volume of acid in L"$

$= 2 m o l / L \times 0.1 L = 0.2 m o l$ $=2 mol"/"Lxx 0.1L=0.2mol$

So the mass of $H C l$ $HCl$ reacted $= 0.2 \times 36.5 g$ $= 0.2xx36.5g$ , where molar mass of $H C l = 36.5 g / m o l$ $HCl =36.5g"/"mol$

Now we know from the law of equivalent proportion that the ratio of masses of two reacting substances for neutralization is equal to the ratio of their equivalent masses.

Hence we can say

$\frac{Equivalent mass of metal carboate}{Equivalent mas of H C l} = \frac{Reacting mass of metal carboate}{Equivalent mas of H C l}$ $"Equivalent mass of metal carboate" /("Equivalent mas of "HCl)="Reacting mass of metal carboate" /("Equivalent mas of "HCl)$

$\Rightarrow \frac{Equivalent mass of metal (x=?)+ equivalent mass of C O_{3}^{2 -}}{Equivalent mas of H C l} = \frac{0.2 \times 36.5}{36.5}$ $=>("Equivalent mass of metal (x=?)+ equivalent mass of "CO_3^(2-))/("Equivalent mas of "HCl)=(0.2xx36.5)/36.5$

$\Rightarrow \frac{x + 30 g/equiv}{36.5 g/equiv} = \frac{10 g}{0.2 \times 36.5 g}$ $=>(x + 30"g/equiv")/(36.5"g/equiv")=(10g)/(0.2xx36.5g)$

$\Rightarrow x = 20 g/equiv$ $=>color(red)(x=20"g/equiv")$

Please note
Equivalent mass of $C O_{3}^{2 -} = \frac{formula mass of C O_{3}^{2 -}}{valency}$ $CO_3^(2-)= (" formula mass of " CO_3^(2-))/"valency "$

$= \frac{12 + 3 \times 16}{2} = 30 g/equiv$ $=(12+3xx16)/2=30"g/equiv"$

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A $100 \cdot m L$ $100mL$ volume of $2.0 \cdot m o l \cdot L^{- 1}$ $2.0mol*L^-1$ hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

3 Answers

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A 100⋅mL100*mL volume of 2.0⋅mol⋅L−12.0*mol*L^-1 hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

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A $100 \cdot m L$ $100mL$ volume of $2.0 \cdot m o l \cdot L^{- 1}$ $2.0mol*L^-1$ hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?