All categories

3 years ago

A mixture of caco3 and (nh4)2co3 is 60.7 % co3 by mass. part a find the mass percent of caco3 in the mixture.

Chemistry

1 answer:

blagie [28]3 years ago

7 0

The mixture contains:

CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass

Let, the total mass = 100 grams

Mass of CaCO3 = x grams

Mass of (NH4)2CO3 = y grams

Thus, x + y = 100 ------------(1)

Mass of CO3 = 60.7% = 60.7 g

Molar mass of CO3 = 60 g/mol

Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles

The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,

moles CaCO3 + moles (NH4)2CO3 = 1.012

mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012

x/100 + y/96 = 1.012---------(2)

based on equation 1 we can write: y = 100-x

x/100 + (100-x)/96 = 1.012

x = 71.2 g

Mass of CaCO3 = 71.2 g

You might be interested in

In a titration of 0.5 M HCl and an unknown concentration NaOH, 25.0 ml of NaOH was required to completely neutralize 25.0 ml of

baherus [9]

When an acid is neutralized by a base, that means moles of H+ = moles of OH-

moles of H+ = 0.5 M * 0.025 L HCl = 0.0125 moles H+

moles of OH- should be equal to 0.0125 moles, so
0.0125 moles = (x) * 0.025 L NaOH

x is the concentration of NaOH, which we want to find.

x = 0.5 M

The correct answer is C) 0.5 M.

6 0

3 years ago

4.41 g of propane gas (C3H8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below

gayaneshka [121]

Answer : The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

$q_{rxn}=-q_{cal}$

$q_{cal}=c_{cal}\times \Delta T$

where,

$q_{rxn}$ = heat released by the reaction = ?

$q_{cal}$ = heat absorbed by the calorimeter

$c_{cal}$ = specific heat of calorimeter = $97.1kJ/^oC=97100J/^oC$

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=(27.282-25.000)=2.282^oC$

Now put all the given values in the above formula, we get:

$q_{cal}=(97100J/^oC)\times (2.282^oC)$

$q_{cal}=221582.2J=221.6kJ$

As, $q_{rxn}=-q_{cal}$

So, $q_{rxn}=-221.6kJ$

Thus, the heat of the reaction is -221.6 kJ

6 0

3 years ago

Which technique would be best for separating sand and water?

ipn [44]

Answer:

A. filtration

Hope it helps

3 0

3 years ago

If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol

maks197457 [2]

Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.

8 0

2 years ago

Is mud a heterogeneous or homogeneous mixture?

Lemur [1.5K]

Answer:

Heterogenous

Explanation:

Mud is a mixture of water and soil

8 0

2 years ago

Read 2 more answers