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Sergeeva-Olga [200]
3 years ago
8

A 2.17 gm sample barium reacted completely with water what is the equation for the reaction how many milliliters of dry H2 evole

d at 21degree celcius and 748mm of Hg ?
Chemistry
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

400 mL

Explanation:

Given data:

Mass of barium = 2.17 g

Pressure = 748 mmHg (748/760 = 0.98 atm)

Temperature = 21 °C ( 273+ 21 = 294k)

Milliliters of H₂ evolved = ?

Solution:

chemical equation:

Ba + 2H₂O →  Ba(OH)₂ + H₂

Number of moles of barium:

Number of moles = mass/ molar mass

Number of moles = 2.17 g / 137.327 g/mol

Number of moles = 0.016 mol

Now we  will compare the moles of barium with H₂.

                       Ba        :       H₂

                         1         :         1

                  0.016        :     0.016

Milliliters of H₂:

PV = nRT

V = nRT/P

V = 0.016 mol ×  0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm

V = 0.39 atm. L/0.98 atm

V = 0.4 L

L to mL

0.4 × 1000 = 400 mL

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How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius
elixir [45]

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

<u>Step 1: </u>Data given

mass of water = 300 grams

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<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

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There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

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3 years ago
Fe2O3 + CO --&gt; Fe + CO2
satela [25.4K]

Answer:

The answer to your question is

1.-Fe₂O₃

2.- 280 g

3.- 330 g

Explanation:

Data

mass of CO = 224 g

mass of Fe₂O₃ = 400 g

mass of Fe = ?

mass of CO₂

Balanced chemical reaction

                       Fe₂O₃   + 3CO    ⇒  2Fe  +   3CO₂

1.- Calculate the molar mass of Fe₂O₃ and CO

Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g

CO = 12 + 16 = 28 g

2.- Calculate the proportions

theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90

experimental proportion Fe₂O₃ / CO = 400/224 = 1.78

As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.

3.-     160 g of Fe₂O₃  --------------- 2(56) g of Fe

         400 g of Fe₂O₃ ---------------  x

         x = (400 x 112) / 160

        x = 280 g of Fe

4.-      160 g of Fe₂O₃  --------------- 3(44) g of CO₂

          400 g of Fe₂O₃  --------------  x

          x = (400 x 132)/160

         x = 330 gr

3 0
3 years ago
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