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ira [324]
3 years ago
9

Is the following equation balanced? 2Na + 2H2O = 2NaOH + H2 Yes No

Chemistry
2 answers:
Vika [28.1K]3 years ago
8 0
Yes there is an equal amount of atoms on either side
Oksi-84 [34.3K]3 years ago
4 0

Answer:

Yes

Explanation:

2Na + 2H2O → 2NaOH + H2

A chemical equation is balanced if the number of atom of elements on the reactant side is equal to the number of atom of elements on the product sides.

According to the equation above the chemical equation is balanced .From the equation 2 moles of sodium reacted with 2 moles of water to form 2 moles of sodium hydroxide and hydrogen gas.

On the reactant side(left side), the sodium has 2 atom of sodium and the water molecule has 4 atom of hydrogen and 2 atom of oxygen.

On the product side(product side), the sodium hydroxide compound have 2 atom of sodium , 2 atoms of oxygen and 2 atoms of hydrogen . The hydrogen gas has 2 atoms of hydrogen. Adding the hydrogen atom on the product side will give you 4 atoms of hydrogen.

From the illustration above you could notice the equation is balanced.

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6 0
3 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
When aluminum is mixed with iron ii oxide iron metal and aluminum oxide?
Alexandra [31]
Are produced along with a large quantitu of heat
6 0
3 years ago
How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

4 0
3 years ago
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