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3 years ago

Sodium chloride reacts with 0.10 moles of aluminum nitrate in water soultion. How many grams of sodium chloride are required?

Chemistry

1 answer:

Vladimir79 [104]3 years ago

6 0

Answer:

The answer to your question is 0.3 moles of NaCl

Explanation:

Data

NaCl = ?

Al(NO₃)₃ = 0.1 moles

Balanced reaction

Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃

Now, use proportions to solve the problem

1 mol of Al(NO₃)₃ ------------ 3 moles of NaCl

0.1 moles of Al(NO₃)₃ -------- x

x = (0.1 x 3) / 1

x = 0.3 moles of NaCl

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A student claims that if she wanted to make a solution quickly, she should use small pellets instead of powder along with heatin

vova2212 [387]

Answer: Yes we agree with the student's claim.

Explanation:

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3 years ago

A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t

erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl

This reaction is the inverted reaction given, so

ΔH = +157 kJ

4 0

3 years ago

What is the mass of 4.99×1021 platinum atoms?

charle [14.2K]

Answer:

$\boxed {\boxed {\sf 1.62 \ g \ Pt}}$

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

$\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

$4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

Flip the ratio so the units of atoms of platinum cancel.

$4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}$

$4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}$

$\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt$

Divide.

$0.008286283627 \ mol \ Pt$

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

$\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

Multiply by the number of moles we just calculated.

$0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

The units of moles of platinum cancel.

$0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }$

$0.008286283627* {195.08 \ g \ Pt}$

$1.61648821\ g \ Pt$

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

$1.62 \ g \ Pt$

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

8 0

3 years ago