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Paha777 [63]
2 years ago
15

Robert places a piece of magnesium metal (Mg) into a test tube of hydrochloric acid (HCl). The magnesium begins to bubble and fi

zz, releasing hydrogen gas (H2) and the test tube gets warm. Which equation represents the reaction Robert observed?
Chemistry
2 answers:
olchik [2.2K]2 years ago
8 0

Answer:

See below.

Explanation:

The balanced equation is

Mg(s) + 2HCl ---->   MgCl2 + H2(g)

ohaa [14]2 years ago
5 0

Answer:

Mg_{(s)} + 2HCl_{(aq)} ---> MgCl_2_{(aq)} + H_2_{(g)}

Explanation:

The reaction is a simple replacement reaction in which hydrogen gas is liberated when it gets replaced by Magnesium in aqueous solution of hydrochloric acid.

<em>1 mole of Mg reacts with 2 moles of HCl to give 1 mole MgCl_2 and 1 mole H_2 gas. The balanced equation for the reaction Robert observed is as follows:</em>

 Mg_{(s)} + 2HCl_{(aq)} ---> MgCl_2_{(aq)} + H_2_{(g)}

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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
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Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
2 years ago
The correct answer(reported to the proper number of significant figures)to the following is
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Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

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