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BaLLatris [955]
3 years ago
12

"how many moles of calcium atoms are in 2.5 moles of calcium carbonate caco3"

Chemistry
1 answer:
Montano1993 [528]3 years ago
5 0

Answer: -

2.5 moles of calcium atoms are in 2.5 moles of calcium carbonate CaCO₃

Explanation: -

In order to solve such types of problems, the first step would be to write the chemical formula of the compound.

The chemical formula of calcium carbonate = CaCO₃

The chemical symbol of Calcium is Ca.

From the formula of calcium carbonate we can see that

1 mole of CaCO₃ has 1 mole of Ca

2.5 mole of CaCO₃ has \frac{1 mole Ca }{1 mole CaCO₃} x 2.5 mole CaCO₃

= 2.5 mol of Ca.

∴2.5 moles of calcium atoms are in 2.5 moles of calcium carbonate CaCO₃

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Jeff used an equal arm balance to weigh a 4.312 g sample of sodium chloride. Which of these measurements made by Jeff is the mos
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Answer:

4.3.

Explanation:

  • Measurements that are close to the known value are said to be accurate, whereas measurements that are close to each other are said to be precise.
  • The most closest measurement to the known value is 4.3 g.
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Which statement is true at STP? (The atomic mass of Zn is 65.39 u.)
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Zinc is a metal. At STP, it exists as solid and is stable as it is. It is an important mineral and is used in many applications like in food, metal and drugs. Zinc can be found in the Earth's crust and also it is present in small amounts in some food.
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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
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Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

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