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3 years ago

A titration is performed on a 25.0 mL sample of calcium hydroxide. A volume of

Chemistry

1 answer:

slamgirl [31]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf 0.28 \ M}}$

Explanation:

Titration is a method used to determine the concentration of a substance. The formula for this is:

$M_AV_A= M_BV_B$

Where M is the molarity of the acid or base and V is the volume of the acid or base.

We know that 46.0 milliliters of a 0.15 molar solution of nitric acid wereused in the titration. 25.0 milliliters of solution of calcium hydroxide of unknown molarity were also used.

$\bullet \ M_A= 0.15 \ M \\\bullet \ V_A= 46.0 \ mL\\\bullet \ V_B= 25.0 \ mL$

Substitute these known values into the formula.

$0.15 \ M * 46.0 \ mL= M_B * 25.0 \ mL$

We are solving for the molarity of the base, so we must isolate the variable $M_B$ . It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

$\frac {0.15 \ M * 46.0 \ mL}{25.0 \ mL}= \frac{M_B * 25.0 \ mL}{25.0 \ mL}$

$\frac {0.15 \ M * 46.0 \ mL}{25.0 \ mL}= M_B$

The units of milliliters (mL) cancel.

$\frac {0.15 \ M * 46.0 }{25.0}=M_B$

$\frac {6.9}{25.0 } \ M =M_B$

$0.276 \ M = M_B$

The original measurements have 2 or 3 significant figures. We always round our answer to the least number of sig figs, which is 2. For the number we calculated, that is the hundredths place. The 6 in the thousandths place tells us to round the 7 up to an 8.

$0.28 \ M \approx M_B$

The concentration of calcium hydroxide is approximately <u>0.28 M.</u>

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) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of c

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The Balanced Chemical Equation is as follow;

4 KO₂ + 2 CO₂   → 2 K₂CO₃ + 3 O₂

First find out the Limiting Reagent,
According to equation,

284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
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Solving for X,
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X =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
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Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g

X =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield = 13.57 / 21.8 × 100

%age Yield = 62.24 %

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