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1 year ago

9

A 1.225 g sample mixture of lithium hydrogen carbonate is decomposed by heating to produce 0.660 g lithium carbonate. Calculate

the theoretical yield and percent yield of Li2CO3

1 answer:

Eva8 [605]1 year ago

4 0

Answer:

The theoretical yield of lithium carbonate = 0.6659 g.

The percent yield = 99.11%.

Explanation:

Lithium hydrogen carbonate is decomposed by heating according to:

<em>2LiHCO₃(s) → Li₂CO₃(s) + H₂CO₃(g).</em>

<em>It is clear that 2.0 moles of lithium hydrogen carbonate produce 1.0 mole of lithium carbonate.</em>

<em />

We need to calculate the no. of moles of (1.225 g) lithium hydrogen carbonate decomposed:

no. of moles of lithium hydrogen carbonate = mass/molar mass = (1.225 g)/(67.96 g/mol) = 0.018 mol.

0.02 mol of lithium hydrogen carbonate is decomposed to produce (0.02 mo /2 = 0.009 mol) of lithium carbonate.

<em>∴ The theoretical yield of lithium carbonate = no. of moles x molar mass</em> = (0.009 mol)(73.89 g/mol) = <em>0.6659 g.</em>

<em>∵ The percent yield = (actual yield/theoretical yield) x 100 = (0.660 g / 0.6659 g) x 100 = 99.11%.</em>

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SOVA2 [1]

Answer:

Mass = 84.82 g

Explanation:

Given data:

Number of molecules of CaSO₄ = 3.75× 10²³

Mass in gram = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

3.75× 10²³ molecule × 1 mol / 6.022 × 10²³ molecules

0.623 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.623 mol × 136.14 g/mol

Mass = 84.82 g

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1 year ago

What volume of 0.500 M HNO₃(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

USPshnik [31]

Answer:

v = 500 milliliters

Explanation:

$HNO_{3}$ ⇒ $H^{+} + NO^{3-}$

$KOH$ ⇒ $K^{+} + OH^{-}$

1 $H^{+}$ to 1 $OH^{-}$

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2 years ago

An ice cube with a volume of 45.0ml and a density of 0.9000g/cm3 floats in a liquid with a density of 1.36g/ml. what volume of t

Grace [21]

Answer : The volume of the cube submerged in the liquid is, 29.8 mL

Explanation :

First we have to determine the mass of ice.

Formula used :

$\text{Mass of ice}=\text{Density of ice}\times \text{Volume of ice}$

Given:

Density of ice = $0.9000g/cm^3=0.9000g/mL$

Volume of ice = 45.0 mL

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The cube will float when 40.5 g of liquid is displaced.

Now we have to determine the volume of the cube is submerged in the liquid.

$\text{Volume of ice}=\frac{\text{Mass of liquid}}{\text{Density of liquid}}$

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$\text{Volume of ice}=29.8mL$

Thus, the volume of the cube submerged in the liquid is, 29.8 mL

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2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

What is the amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follow

Irina18 [472]

From the equation, we can tell that 1 mol of Al₂S₃ requires 6 moles of water.
The molar ratio is 1/6
Moles of Al₂S₃ present = 20/150.17
= 0.133
Moles of water present = 2/18.02
= 0.111
The moles of Al₂S₃ that will react are:
0.111/6
= 0.0185
The remaining amount:
0.133 - 0.0185
= 0.1145 mol
Or
0.1145 * 150.17
= 17.19 grams

6 0

2 years ago

Read 2 more answers