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3 years ago

CaSO4 + AIP --> Ca3P2 + __AI2(SO4)3

Chemistry

1 answer:

4vir4ik [10]3 years ago

3 0

Answer:

4.89 mol

Explanation:

Step 1: Write the balanced equation

This is a double displacement reaction.

3 CaSO₄ + 2 AIP ⇒ Ca₃P₂ + AI₂(SO₄)₃

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of CaSO₄ to AlP is 3:2.

Step 3: Calculate the moles of AlP needed to react with 7.33 moles of CaSO₄

We will use the previously established molar ratio.

7.33 mol CaSO₄ × 2 mol AlP/3 mol CaSO₄ = 4.89 mol AlP

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How many moles of \ce{AgCl}AgClA, g, C, l will be produced from 60.0 \text{ g}60.0 g60, point, 0, start text, space, g, end text

lozanna [386]

Answer:

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Explanation:

How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

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For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

6 0

4 years ago

A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib

rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

Now put all the given values in this expression, we get:

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

7 0

4 years ago

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Rainbow [258]

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3 years ago

Answer Quick

kow [346]

Answer:

20.9%

Explanation:

The percentage by mass of solution is given by dividing the mass of solute in grams by the mass of solution in grams then multiplying it by 100%.

% Mass of solution = mass of solute/mass of solution × 100%

= (27.0 g/ 129.0 g) × 100%

= 20.93%

= 20.9%

7 0

3 years ago

Read 2 more answers

__ CaSO4 + ____AIP --> ____Ca3P2 + ____AI2(SO4)3

CaSO4 + AIP --> Ca3P2 + __AI2(SO4)3