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3 years ago

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Chemistry

1 answer:

agasfer [191]3 years ago

5 0

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02 mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02 mol × 6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

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3 years ago

What volume of 0. 250 m sulfuric acid solution would be needed to react completely with 18. 00 ml of 0. 350 m ammonia solution?

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2 years ago

When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balan

Damm [24]

Answer:

a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) 67.6%

Explanation:

a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).

Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.

In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:

Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.

First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:

1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄

Transforming to mass (mass = moles * molar mass):

331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄

x ------------------------- 12.38

By a simple direct three rule:

162x = 4100.256

x = 25.3 g of Pb(NO₃)₂

This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:

1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)

Transforming to mass:

331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)

11.39 g ------------------- y

By a simple direct three rule:

331.2 y = 3681.248

y = 11.115 g

The yield is the mass formed divided by the stoichiometric result multiplied by 100%:

yield = (7.52/11.115)*100% = 67.6%

7 0

3 years ago