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2 years ago

What is the mass of 2.9 moles of calcium? Explain please

Chemistry

1 answer:

Alik [6]2 years ago

4 0

Answer:

116 g

Explanation:

From the question given above, the following data were obtained:

Number of mole of calcium = 2.9 moles

Mass of calcium =.?

The mole and mass of a substance are related according to the following formula:

Mole = mass / molar mass

With the above formula, we can obtain the mass of calcium. This can be obtained as follow:

Number of mole of calcium = 2.9 moles

Molar mass of calcium = 40 g/mol

Mass of calcium =.?

Mole = mass / molar mass

2.9 = mass of calcium / 40

Cross multiply

Mass of calcium = 2.9 × 40

Mass of calcium = 116 g

Therefore, the mass of 2.9 moles of calcium is 116 g.

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

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Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 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 = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

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

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

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

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

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ 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 = 0.26 moles

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

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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What causes some aqueous solutions to have a low PH?

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