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2 years ago

How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?

Chemistry

1 answer:

zalisa [80]2 years ago

5 0

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

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Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. cons

Dmitriy789 [7]

Lithium sulfate is Li2SO4

Strontium chlorate is Sr(ClO3)2

Strontium sulfate is SrSO4

So the complete balanced chemical reaction for this is:

Li2SO4 (aq) + Sr(ClO3)2 (aq) --> SrSO4 (s) + 2 LiClO3 (aq)

This is a type of double replacement reaction since there is an exchange of ions.

4 0

3 years ago

The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is

kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

Gradient = $\frac{[A]_{2} - [A]_{1} }{t_{2} - t_{1} }$ where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M

= $\frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s}$ = -0.001056 M/s = -1.056×10⁻³ M/s

Threfore k = 1.056×10⁻³ M/s

3 0

3 years ago

Please help! it’s due date is in a few minutes

Licemer1 [7]

Answer:

40'0'

Explanation:

X is smaller than |-40|

7 0

2 years ago

If the air in a room has lower temperature than the air coming out of a floor vent from a forced air heater , heat would flow

jolli1 [7]

To the top. Common knowledge my dude. Heat always rises to the top.

5 0

3 years ago

Read 2 more answers

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

2 years ago