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Ulleksa [173]
3 years ago
10

1) How many grams of CO are needed to react with an excess of Fe2O3 to produce 225.5 gFe? Numbers after symbols are subscripts u

nless there is a ^ to show it is an exponent. ........... Fe2O3(s) +3CO(g) -- > 3 CO2(g) + 2 Fe(s) ...........
2) Solid sodium reacts violently with water producing heat, hydrogen gas and sodium hydroxide. How many molecules of hydrogen gas are produced when 65.4 g of sodium are added to water? Numbers after symbols are subscripts unless there is a ^ to show it is an exponent. ............ 2 Na(s) +2 H2O(l) -- > 2 NaOH(aq) + H2(g) ...........
Chemistry
1 answer:
Mrrafil [7]3 years ago
6 0
1)
Moles = mass / Mr
Mr of Fe = 56
Moles of Fe = 225.5 / 56
Moles of Fe = 4.03

Molar ratio CO : Fe = 3 : 2
Moles of CO = 4.03 x 3/2
Moles of CO = 6.04 moles

Mass of CO = 6.04 x (12 + 16)
Mass of CO = 169.1 grams


2)
Moles of Na = 65.4 / 23
Moles of Na = 2.84

Molar ratio Na : H₂ = 2 : 1
Moles of H₂ = 1.42
Each mole has 6.02 x 10²³ particles

Molecules of Hydrogen = 1.42 x 6.02 x 10²³
Molecules of Hydrogen = 8.55 x 10²³ particles
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A calorimeter contains 35.0 mLmL of water at 15.0 ∘C∘C . When 2.20 gg of XX (a substance with a molar mass of 56.0 g/molg/mol )
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Answer:

ΔH rx = -43.5 kJ / mol

Explanation:

In water, Xdissolves thus:

X(s) + H₂O(l) → X(aq) + H₂O(aq)

It is possible to find the heat in dissolution process using coffee cup calorimeter equation:

Q = -m×C×ΔT

<em>Where Q is heat, m is mass of solution (35.0g -density 1g/mL- + 2.20g = 37.2g), C is specific heat of solution (4.18J/g°C), and ΔT is change in temperature (26.0°C-15.0°C = 11.0°C)</em>

Replacing:

Q = -37.2g×4.18J/g°C×11.0°C

Q = -1710J = -<em>1.71kJ</em>

As enthalpy is the change in heat per mole of reaction, moles of X that reacted were:

2.20g X × (1mol / 56.0g) = <em>0.0393 moles</em>

As heat produced per 0.0393moles was -1.71kJ, heat per mole of X is:

-1.71kJ / 0.0393mol = -<em>43.5 kJ / mol = ΔH rx</em>

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