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olganol [36]
3 years ago
7

What is the mole ratio to correct balance the reaction below? __KBr + __Ca3N2--> __CaBr2 + __K3N

Chemistry
1 answer:
jonny [76]3 years ago
7 0

Explanation:

3CaBr2 + 2K3N → 6KBr + Ca3N2

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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Harman [31]
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y

3y = 1.2
y = 0,4M

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4

C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄

mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol

164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
5 0
3 years ago
Which of the following would most likely happen if water did not form hydrogen bonds?
Oksanka [162]
Water would not expand when it freezes.<span> </span>
8 0
3 years ago
Read 2 more answers
How do you name these compounds?
8090 [49]
1) is right
2) lead(ii) phosphate
3) iron(iii)sulfate
4) lead(ii)oxide
5) lead sulfate
6) copper iodide
7) lead oxide
8 0
3 years ago
Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur
yuradex [85]

<u>Answer:</u> The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For magnesium:</u>

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol

  • <u>For iron(III) chloride:</u>

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = \frac{2}{3}\times 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0
3 years ago
All rocks follow the same pathway though the rock cycle us a false statement. Why?
sammy [17]

False.

This is because there are three different rocks, metamorphic, sedimentary, and igneous. Thus, meaning that there are three different pathways, making this statement false.

Hope this helps!

3 0
3 years ago
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