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PtichkaEL [24]
3 years ago
13

Given a balanced chemical equation below: 3Cu(s) + 2H3PO4 --- > Cu3(PO4)2 + 3H2 How many moles of copper are needed to react

with 5 moles of phosphoric acid?
Chemistry
1 answer:
Harman [31]3 years ago
4 0

Answer:

7.5 moles

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2

From the balanced equation above,

3 moles of Cu reacted with 2 moles of H3PO4.

Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e

Xmol of Cu = (3 x 5)/2

Xmol of Cu = 7.5 moles

Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.

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slamgirl [31]

Answer:

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Explanation:

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5 0
3 years ago
How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

8 0
3 years ago
What is the mass of sodium (Na) in 50 grams of table salt (NaCl)? Show your work.
Margaret [11]

Answer:

19 g

Explanation:

Data Given:

Sodium Chloride (table salt) = 50 g

Amount of sodium (Na) = ?

Solution:

Molecular weight calculation:

NaCl = 23 + 35.5

NaCl = 58.5 g/mol

Mass contributed by Sodium = 23 g

calculate the mole percent composition of sodium (Na) in sodium Chloride.

Since the percentage of compound is 100

So,

Percent of sodium (Na) = 23 / 58.5 x 100

Percent of sodium (Na) = 39.3 %

It means that for ever gram of sodium chloride there is 0.393 g of Na is present.

So,

for the 50 grams of table salt (NaCl) the mass of Na will be

mass of sodium (Na) = 0.393 x 50 g

mass of sodium (Na) = 19 g

8 0
3 years ago
Read 2 more answers
Several science students were about to conduct an experiment on electricity. The teacher passed on the science activity which cl
olganol [36]
C. You should ALWAYS ask the teacher if you don't get something; your friends could be wrong, don't guess it, and NEVER cheat. Hope this helps!!
8 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
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