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3 years ago

Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? (2 points)

Chemistry

2 answers:

vazorg [7]3 years ago

8 0

The answer would be the first option - k

serg [7]3 years ago

8 0

Answer: The correct answer is Potassium (K).

Explanation:

Alkali (Group 1) and Alkaline earth (Group 2) elements react with water to produce metal hydroxide and also releases hydrogen gas.

Reaction of group 1 elements with water:

$2M+2H_2O\rightarrow 2MOH+H_2$

Reaction of group 2 elements with water:

$M+2H_2O\rightarrow M(OH)_2+H_2$

From the given options:

Potassium (K): This metal belongs to Group 1 and readily reacts with water to form potassium hydroxide and hydrogen gas.

$2K+2H_2O\rightarrow 2KOH+H_2$

Tin (Sn): This element belongs to Group 14 and does not readily reacts with water.

Nickel (Ni): This element belongs to Group 10 and does not readily reacts with water.

Manganese (Mn): This element belongs to Group 7 and does not readily reacts with water.

Hence, the correct answer is potassium (K).

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Answer:

Explanation:

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3 years ago

What mass of CO is needed to react completely with55.0 g of Fe2O3(s)+CO(g) yield Fe(s)+CO2(g)?

Katarina [22]

Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Gather all the information in one place with molar masses above the formulas and masses below them.

$M_{r}$ : 159.69 28.01

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass/g: 55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

$\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}$

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

$\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}$

3.Use the molar mass of CO to calculate the mass of CO.

$\text{Mass of CO} = \text{1.033 mol CO} \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}$

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