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2 years ago

How many moles of Mg3(PO4)2 are in 350.00 grams of Mg3(PO4)2?

Chemistry

2 answers:

dusya [7]2 years ago

6 0

Answer:

1.3 moles/ 1.33150727 moles

Explanation:

350g x 1 mol/262.86g = 1.3 moles

jok3333 [9.3K]2 years ago

6 0

The molar mass of Mg3(PO4)2 is 262.855 g/mol

Atomic weight of the parts:

Mg 24.3050 x 3 = 72.92

P 30.973762 x2 = 61.95

O 15.9994 x8 = 127.995

There are 262.855 grams in one mole, so

350.00 g x (262.855g/1 mole) = 1.33 moles Mg3(PO4)2

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Which of the following is true of a heterogeneous mixture?

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Explain the statement - all bases are not alkalis but all alkalis are bases

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A substance that neutralizes an acid is a base and that soluble in water is an alkali. However, all bases are not soluble in water. Thus, All alkali are bases but all bases are not alkali.

Explanation:

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2 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

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Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

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Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

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2 years ago

For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(

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Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

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moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

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Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

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