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3.
CaCO3 + 2HCl --> CaCl₂ + H₂O + CO2
How much 0.80 M HCI would be needed to dissolve a CaCO3 pearl
which weighs 4.0 grams?

4.
3 Fe + 2 Au(NO3)3 --> 3 Fe(NO3)2 + 2 Au.
Throwing some scrap iron in a gold nitrate solution causes the gold
metal to precipitate. How much 0.50 M gold nitrate solution would
react with 224 grams of iron metal?

5. Sea water is about 0.50 M NaCl. To produce Cl2 gas, a company
evaporates sea water, melts the NaCl, and runs electricity through it.
2 NaCl -> 2 Na + Cl2 How many liters of sea water are needed
to fill a tank car with 45800 grams of chlorine gas

6.
H3PO4 + 3 NaOH ----> Na3PO4 + 3 H₂O
If 36.0 mL of H3PO4 react exactly with 80.0 ml. of 0.500 M NaOH,
what is the concentration of the phosphoric acid?
(molarity)

3. .10 L
4. 5.3 L
5. 2600 L
6. .370 M

Answer :

Stoichiometry is used to describe the relationship between the reactants and products in a  chemical equation.

What is stoichiometry?

The term stoichiometry is used to describe the relationship between the reactants and products in a  chemical equation. We shall now apply the concept of stoichiometry to answer these questions.

1) Reaction equation: CaCO3 + 2HCl --> CaCl₂ + H₂O + CO2

Number of moles of 4 g of CaCO3 = 4g/100 g/mol = 0.04 moles

1 mole of CaCO3 reacts with 2 moles of HCl

0.04 moles of CaCO3 reacts with 0.04 moles * 2 moles/1 mole

= 0.08 moles

Since;

number of moles = concentration * volume

volume = number of moles/concentration

volume = 0.08 moles/ 0.80 M = 0.1 L or 100 mL

2) Reaction equation; Fe + 2 Au(NO3)3 --> 3 Fe(NO3)2 + 2 Au.

Number of moles of Fe = 224 g/56 g/mol = 4 moles

1 mole of Fe reacts with 2 moles of Au(NO3)3

4 moles  of Fe reacts with 4 moles  * 2 moles/1 mole = 8 moles

Volume of Au(NO3)3 = 8 moles/0.50 M  = 16 L

3) The reaction equation is; 2 NaCl -> 2 Na + Cl2

Number of moles of Cl2 gas = 45800 g/71 g/mol = 645 moles

Volume of seawater that contains 645 moles of Cl = 645 moles/ 0.50 M

= 1290 L

4) The reaction equation is; H3PO4 + 3 NaOH ----> Na3PO4 + 3 H₂O

From;

Number of moles of NaOH  = 80/1000 * 0.500 M  = 0.04 moles

If 1 mole of H3PO4  reacts with 3 moles of NaOH  

x moles of H3PO4  reacts with 0.04 moles of NaOH  

x = 0.013 moles

Concentration of H3PO4  = 0.013 moles * 1000/36 = 0.370 M

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