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Scorpion4ik [409]
2 years ago
15

If you start with 30.0 grams of sodium fluoride, how many grams of magnesium fluoride will be produced?

Chemistry
1 answer:
Cloud [144]2 years ago
7 0

Answer:

22.2 g of MgF₂

Explanation:

We star from the reaction:

NaF → sodium fluoride

Mg → Magnesium

2NaF  +  Mg →  MgF₂  +  2Na

2 moles of sodium fluoride react to 1 mol of Mg in order to produce 1 mol of magnesium fluoride and 2 moles of sodium.

We convert mass of the reactant to moles:

30 g . 1mol / 41.98 g = 0.715 mol

We assume that Mg is in excess

As ratio is 2:1, if we have 0.715 moles of NaF we may produce the half of moles, of MgF₂

0.715 mol /2 = 0.357 moles of MgF₂

We convert moles to mass: 0.357 mol . 62.30g /mol = 22.2 g

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2Al₂O₃ = 4Al + 3O₂

M(A₂O₃)=101.96 g/mol
m(Al₂O₃)=250 g

n(O₂)=3m(Al₂O₃)/{2M(Al₂O₃)}

n(O₂)=3*250/{2*101.96}=3.678 mol
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09. Where does Mitosis take place? I
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What is the percent lithium in lithium nitrate?
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3 years ago
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
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