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Mashutka [201]
3 years ago
11

Calculate the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid according to the balanced equat

ion: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
Chemistry
1 answer:
Alex3 years ago
6 0

6.069 grams is the  mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid.

Explanation:

Balanced equation for the reaction:

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

data given:

mass of aluminum = 27 grams

atomic mass of one mole of aluminum = 26.89 grams/mole

formula to calculate number of moles:

number of moles = \frac{mass}{atomic mass of one mole}

number of moles = \frac{27}{26.89}

                              = 1.004 moles of aluminum will react

from the balanced equation:

2 moles of Al reacted to form 3 moles of H2

1.004 moles of Al will produce x moles of H2

\frac{3}{2} = \frac{x}{1.004}

x = 3.012 moles of H2 will be formed.

mass will be calculated as number of moles multiplied by atomic weight

mass of 3.012 moles of hydrogen ?(atomic weight of one mole H2 = 2.015 grams)

= 3.012 x 2.015

= 6.069 grams of H2 will be formed.

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A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams
jek_recluse [69]

The mass of methanol produced is 8.0 g.

We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

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

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01  2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO

Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO  × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

3 0
3 years ago
A solution was prepared by mixing 50.0 g
frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0
3 years ago
A student had a sample of pure water, and added an unknown substance to it. The student noticed that the hydroxide ion concentra
steposvetlana [31]
The answer is A it's a basic because once you add another substance to a neutral it either becomes acidic or basic. this one becomes basic because the hydroxide ion concentrate increased.
6 0
3 years ago
Read 2 more answers
60 POINTS! PLEASE ANSWER!
Mumz [18]
I would have helped but I didn’t understand it sorry that I didn’t answer :(
6 0
2 years ago
PLEASE TELL ME IF THIS IS RIGHT CLASS IN % MINUTES
alexdok [17]

Answer:

I think that is pretty good and if you get it wrong you showed your work so your teacher can see you're thinking. That is defenetly gonna get you a score.

Explanation:

7 0
3 years ago
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