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2 years ago

If 8.23 g of magnesium chloride react completely with sodium phosphate, how many grams of magnesium phosphateare produced

Chemistry

1 answer:

steposvetlana [31]2 years ago

8 0

Answer:

The correct answer is 7.57 grams of magnesium phosphate.

Explanation:

Based on the given question, the chemical reaction taking place is:

2Na₃PO₄ (aq) + 3MgCl₂ (aq) ⇒ Mg₃(PO₄)2 (s) + 6NaCl (aq)

From the given reaction, it is evident that two moles of sodium phosphate reacts with three moles of magnesium chloride to produce one mole of magnesium phosphate.

Based on the given information, 8.23 grams of magnesium chloride reacts completely with sodium phosphate, therefore, magnesium chloride in the given case is the limiting reagent.

In the given case, 3 moles of magnesium chloride produce 1 mole of magnesium phosphate. Therefore, 1 mole of magnesium chloride will produce 1/3 mole of magnesium phosphate.

The molecular mass of magnesium chloride is 95.21 grams per mole. So, 1 mole of magnesium chloride is equivalent to 95.21 grams of magnesium chloride.

On the other hand, the molecular mass of magnesium phosphate is 262.85 grams per mole. Therefore, 1 mole of magnesium phosphate is equal to 262.85 grams of magnesium phosphate.

As seen earlier that 1 mole of magnesium chloride = 1/3 moles of magnesium phosphate. So,

95.21 grams of magnesium chloride = 1/3 × 262.85 grams of magnesium phosphate

= 262.85 / 3 grams of magnesium phosphate

1 gram of magnesium chloride = 262.85 / 3 × 95.21 grams of magnesium phosphate

8.23 grams of magnesium chloride = 262.85 / 3 × 95.21 × 8.23 grams of magnesium phosphate

= 7.57 grams of magnesium phosphate

Hence, when 8.23 grams of magnesium chloride when reacts completely with sodium phosphate, it produces 7.57 grams of magnesium phosphate.

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3 years ago

Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans

loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

Answer: The pH of the solution is 10.4

Explanation:

We are given:

Molarity of sodium hypochlorite = 0.39 M

$pK_a$ of HClO = 7.50

We know that:

$pK_a=-\log K_a$

$K_a$ of HClO = $10^{-7.50}=3.16\times 10^{-8}$

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant = $3.16\times 10^{-8}$

$K_b$ = Base dissociation constant

Putting values in above equation, we get:

$10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}$

The chemical equation for the reaction of hypochlorite ion with water follows:

$ClO^-+H_2O\rightarrow HClO+OH^-$

Initial: 0.39

At eqllm: 0.39-x x x

The expression of $K_b$ for above equation follows:

$K_b=\frac{[HClO][OH^-]}{[ClO^-]}$

Putting values in above equation, we get:

$3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035$

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.00035M$

Putting values in above equation, we get:

$pOH=-\log (0.00035)=3.6$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.6=10.4$

Hence, the pH of the solution is 10.4

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