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vichka [17]
3 years ago
9

When the reaction h3po4(aq) ca(oh)2(aq)→ occurs what is the formula for the salt formed?

Chemistry
2 answers:
Georgia [21]3 years ago
7 0
2H3PO4 + 3Ca(OH)2 ➡ Ca3(PO4)2 + 6H2O.

Hence salt formed is Ca3(PO4)2.
Hope this helps, have a nice day!
ladessa [460]3 years ago
7 0

Answer:

The formula for the salt formed is Ca(HPO4)

Explanation:

When an acid reacts with a base, there is a salt formation. Sometimes the replacement of the hidrogens in the acid by the cation of the base is complete, and sometimes there is a partial replacement,as in the presente case. Here, the cation of the base is Ca2+ and it will reacts with the anion  HPO4- of the acid.

H3PO4 + Ca(OH)2 ⇒ Ca(HPO4) + 2H2O

As you can see, water is also formed as a product of the reaction.

Summarizing, the formula for the salt formed is Ca(HPO4)

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Alfred Wegener came up with the idea 
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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine
Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

  • Firstly, we should write the balanced equation of the reaction:

<em>N₂ + 3H₂ → 2NH₃.</em>

<em>1) To determine the limiting reactant of the reaction:</em>

  • From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
  • This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
  • We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

  • From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>

  • As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
  • Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
  • The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
  • ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0
2 years ago
a 20ml sample of hcl was titrated with the 0.0220 M Naoh. to reach the endpoint required 23.72 mL of the NaOh. Calculate the mol
Y_Kistochka [10]

Answer:

Molarity of HCl=0.026092M

Explanation:

The equation for the reaction is;

HCl + NaOH ⇒ NaCl + H2O

Using the formular, \frac{C_{A}V_{A}}{C_{B}V_{B} }=\frac{nA}{nB}    ..........equ1

whereC_{A} is the concentration of Acid,

          V_{A} is volume of acid

          C_{B} is concentration of the base

          V_{B} is volume of the base

          nA is the number of moles of Acid

          nB is number of moles of base

nA = 1,    nB=1 , V_{A}=20ml, C_{B}=0.022M, V_{B}=23.72mL

we will input these values into equation1 to solve for C_{A}

\frac{C_{A}*20}{0.022*23.72}=\frac{1}{1}

C_{A}*20=0.022*23.72

C_{A}=0.026092M

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“The chemical formula in the structure of phosphorus is shown below. This represents a(n)....”
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Answer:

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Explanation:

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What is the concentration of a phosphoric acid solution of a 25.00 mL sample if the acid requires 42.24 mL of 0.135 M NaOH for n
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0.0760 m

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finding the moles of NaOH which will be <span>5.702 E -3 m
</span>
next find the moles of H3PO4 which will be <span>1.90 E -3 m</span><span>
calulcate </span>25 ml sample molarity = 0.07603 m, just put 0.0760<span>

</span>
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