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4 years ago

Convert mass to moles for both reactants. (round to 2 significant figures.)

Chemistry

2 answers:

Maslowich4 years ago

4 0

Answer:

Explanation:

Given parameters:

Mass of CuCl₂ = 2.50g

Mass of Al = 0.50g

Unknown:

Number of moles of CuCl₂ and Al = ?

Solution:

To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CuCl₂ = 63.6 + 2(35.5) = 134.5g/mole

Molar mass of Al = 26.98g/mole

Number of moles of CuCl₂ = $\frac{2.5}{134.5}$ = 0.019moles

Number of moles of Al = $\frac{0.5}{26.98}$ = 0.019moles

a_sh-v [17]4 years ago

4 0

Answer:

2.50g CuCl2 equals 0.019 moles

0.25g Al equals 0.0093 or .0093 moles

Explanation:

Edge2020

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Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.

natima [27]

The question is incomplete, here is the complete question:

Consider the reaction $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initially, $[NH_3(g)]=[O_2(g)]$ = 3.60 M; at equilibrium $[N_2O_4(g)]=0.60M$ . Calculate the equilibrium concentration for $NH_3$

<u>Answer:</u> The equilibrium concentration of ammonia is 2.8 M

<u>Explanation:</u>

We are given:

Initial concentration of $[NH_3(g)]$ = 3.60 M

Initial concentration of $[O_2(g)]$ = 3.60 M

For the given chemical equation:

$4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initial: 3.60 3.60

At eqllm: 3.60-4x 3.60-7x 2x 6x

We are given:

Equilibrium concentration of $[N_2O_4(g)]$ = 0.60 M

Evaluating the value of 'x'

$\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2$

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Why must oxidation be accompanied by a reduction?

Crazy boy [7]

Answer:

Explanation:

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A sample of P2Cl5 contains 179 g of phosphorous, how many grams of chlorine are present? Please show step by step thx!

natima [27]

Answer:

About 512 g.

Explanation:

We are given a sample of P₂Cl₅ that contains 179 grams of phosphorus, and we want to determine the grams of chlroine that is present.

Thus, we can convert from grams of phosphorus to moles of phosphorus, moles of phosphorus to moles of chlorine, and moles of chlorine to grams of chlorine.

From the formula, there are two moles of P for every five moles of Cl. The molecular weights of P and Cl are 30.97 g/mol and 35.45 g/mol, respectively. Hence:

$\displaystyle 179\text{ g P} \cdot \frac{1\text{ mol P}}{30.97\text{ g P}} \cdot \frac{5\text{ mol Cl}}{2\text{ mol P}} \cdot \frac{35.45\text{ g Cl}}{1\text{ mol Cl}} = 512\text{ g Cl}$

In conclusion, there is about 512 grams of chlorine present in the sample.

Alternatively, we can mass percentages. The mass percent of phosphorus in P₂Cl₅ is:

$\displaystyle \% \text{P} = \frac{2(30.97)}{5(35.45) + 2(30.97)} = 25.90\%$

Because there are 179 grams of phosphorus, the total amount of sample present is:

$\displaystyle \begin{aligned} 25.90\% \cdot m_T & = 179\text{ g P} \\ \\ m_T & = 691.1 \text{ g}\end{aligned}$

Therefore, the amount of chlorine present is 691.1 g - 179 g, or about 512 g, in agreement with our above answer.

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