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olya-2409 [2.1K]

3 years ago

14

How many moles of copper (Cu) are in 65.8 g Cu?

2 answers:

DedPeter [7]3 years ago

7 0

How many moles of copper (Cu) are in 65.8 g Cu?

1.04 mol Cu CORRECT ANSWER

10.9 mol Cu

41.7 mol Cu

63.5 mol Cu

I took the test and A) was correct

Veseljchak [2.6K]3 years ago

5 0

<h2><u>Answer:</u></h2>

The correct answer is A) 1.04 mol Cu

{65.8 g / 63.55 g/mol}

= 1.04 mol Cu

Explanation:

In 63.55 g of copper metal there are 1 m o l of C u atoms. By dividing the mass of Cu and molar mass, we can easily get the number of moles.

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Consider the relationship (y+3)2 = b/(x-2), where y and x are variables and bis a constant. On rectangular coordinate paper, wha

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(-1) is the slope of a graph of In(y+3) on the vertical axis versus In(x-2) on the horizontal axis.

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Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

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Answer: 1) Maximum mass of ammonia 198.57g

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g

Explanation:

In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

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3 years ago