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Chemistry

1 answer:

Westkost [7]2 years ago

8 0

Answer:

mNaNO3 =765g

Explanation:

First, we write the balanced chemical equation representing the chemical reaction that happened between aluminum nitrate and sodium chloride.

Balanced chemical equation:

AL(NO3)3+3NaCl ⟶ 3 NaNO3+AlCl3

According to the equation, each mole of aluminum nitrate requires three moles of sodium chloride. Thus, the required number of moles of sodium chloride is

4 Mol ⋅ 3 = 12mol

Based on the data provided in the table, there were 9 moles of sodium chloride used in the reaction, which was not enough for the entirety of aluminum nitrate to react. So, sodium chloride must have been the limiting reactant.

Therefore, we use the number of moles (n) of sodium chloride to calculate the number of moles of sodium nitrate, which has a 1:1 ratio with sodium chloride.

Number of moles sodium nitrate:

nNaNO3=nNaCl

nNaNO3 = 9 mol

We can also calculate the mass (m) of sodium nitrate that was produced by multiplying its number of moles by its molar mass (MM), 85.00g/mol.

Mass of sodium nitrate produced:

mNaNO3 = nNaNO3 ⋅ MMNaNO3

mNaNO3 = 9 mol ⋅ 85.00 g/mol

mNaNO3 =765g

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$