Ask question

Ask question

All categories

3 years ago

8

What is the maxiumum volume of H2 gas at STP that can be produced when 14.0 g of CaH2 and 28.0 g H2O react?

1 answer:

astraxan [27]3 years ago

6 0

You are given the mass of CaH2 which is 14 grams and H2O which is 28 grams. You are required to find the maximum volume of H2 gas at STP. The balanced chemical reaction is CaH2 + 2H2O -> Ca(OH)2 + H2. Note that for every one mole of CaH2, 2 moles of H2O is needed to completely react and produce Ca(OH)2 + H2. So the CaH2 and H2O are at equal proportions. At STP, temperature is at 0°C(273K) and pressure at 1 atm. The molar mass of CaH2 is 42 grams per mole.

14g CaH2(1mol CaH2/42 grams CaH2)(1mol H2/1mol CaH2) = 0.333 moles H2

The ideal gas equation is PV=nRT, with R(gas constant) = 0.08206 L-atm/mol-K. Get the equation for volume, we have

V=nRT/P
V=(0.333molH2)( 0.08206 L-atm/mol-K)(273K)/1atm
<u>V=7.47L</u>

You might be interested in

Could the Periodic Table be arranged differently?

coldgirl [10]

Answer:

No because it is stayed that way and you can't define them differently.

7 0

3 years ago

Read 2 more answers

What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm

Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

$n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}$

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = $5\times 196.96$ g

4.99 moles of gold contain = 984.8 g

Mass of ${3.01\times 10^{24}}$ atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = $19.32 g/cm^{3}$

Volume of the cuboid = $length\times breadth\times height$

Volume of the gold bar = $6.00\times 4.25\times 2.00$

Volume of the gold bar = 51 $cm^{3}$

Using formula,

$Density = \frac{mass}{Volume}$

$Mass = Density\times Volume$

$Mass = 19.32 \times 51$

Mass = 985.32 g

So, A gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

3 0

3 years ago

What is the hardest metal?

marissa [1.9K]

Answer:

i believe it is titanium.

Explanation:

4 0

3 years ago

Read 2 more answers

A solution composed of 30.6g NH3 in 81.3g of H20. Calculate the mole fraction for NH3 and H20

irina [24]

Answer:

0.286 moles

Explanation:

i hope this is helpful for you

4 0

2 years ago

Read 2 more answers

The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti

lisabon 2012 [21]

Answer:

Option E, Half life = $2.96\times 10^3\ s$

Explanation:

For a first order reaction, rate constant and half-life is related as:

$t_{1/2}=\frac{0.693}{k}$

Where,

$t_{1/2}$ = Half life

k = Rate constant

Rate constant given = $2.34\times 10^{-4}\ s^{-1}$

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{2.34 \times 10^{-4}}=2.96\times 10^3\ s$

So, the correct option is option E.

4 0

4 years ago