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1 year ago

(You do) If you have 47.2 mol of Na available, along with an excess of Cl₂, how many grams of NaCl can you produce?

Chemistry

1 answer:

IrinaVladis [17]1 year ago

5 0

Answer:

2,760 grams NaCl

Explanation:

To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.

2 Na + Cl₂ --> 2 NaCl

Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl) = 58.44 g/mol

47.2 moles Na 2 moles NaCl 58.44 grams

---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl

= 2,758.368 grams NaCl

= 2,760 grams NaCl

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

aleksandr82 [10.1K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 13.83 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.300 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.300M=\frac{\text{Moles of NaI}}{0.200L}\\\\\text{Moles of NaI}=(0.300mol/L\times 0.200L)=0.06mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.06 moles of NaI will produce = $\frac{1}{2}\times 0.06=0.03mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

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

Moles of lead (II) iodide = 0.03 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.03mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.03mol\times 461.1g/mol)=13.83g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 13.83 grams

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3 years ago

How many millimeters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3

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So we need 0.276 moles of HCl to react. Your concentration is given in moles/liter so 0.276/1.58 = 0.174 liters needed or 174 milliliters

8 0

3 years ago

Read 2 more answers

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kow [346]

Answer : The final concentration of $N_2O_5$ is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $5.89\times 10^{-3}\text{ min}^{-1}$

t = time passed by the sample = 3.5 min

a = initial concentration of the reactant = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

$3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}$

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Thus, the final concentration of $N_2O_5$ is, 2.9 M

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3 years ago

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Answer:

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Explanation:

1. Word equation:

mercury(II) oxide → mercury + oxygen

2. Balanced molecular equation:

2HgO → 2Hg + O₂(g)

3. Mole ratio

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

4. Calculate the number of moles of O₂(g)

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

5. Calculate the number of moles of HgO

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

6. Convert to mass

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3 years ago

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Ne4ueva [31]

Answer:

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Explanation:

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