How many $M g^{2 +}$ $Mg^(2+)$ ions are present in 3.00 moles of $M g C l_{2}$ $MgCl_2$ ?

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How many $M g^{2 +}$ $Mg^(2+)$ ions are present in 3.00 moles of $M g C l_{2}$ $MgCl_2$ ?

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Answer 1 · 2018-08-06T23:46:05

$1.81 x 10^{24} M g^{2 +} ions$ $1.81" x "10^24 Mg^(2+) "ions"$

Explanation:

Step 1:
From the formula $M g C l_{2}$ $MgCl_2$ , we know that it's made from 1 $M g^{2 +}$ $Mg^(2+)$ and 2 $C l^{-}$ $Cl^-$ ions

In 1 mol of $M g C l_{2}$ $MgCl_2$ , there's 1 mol of $M g^{2 +}$ $Mg^(2+)$

We can use this relationship to calculate the moles of $M g^{2 +} ions$ $Mg^(2+) "ions"$ from the given 3.00 moles of $M g C l_{2}$ $MgCl_2$ .

Step 2:
Once we find the moles of $M g^{2 +}$ $Mg^(2+)$ , we can find the number of $M g^{2 +} ions$ $Mg^(2+) "ions"$ using Avogadro's number.

1 mol $M g^{2 +}$ $Mg^(2+)$ has $6.022$ $6.022$ x $10^{23}$ $10^23$ $M g^{2 +} ions$ $Mg^(2+) "ions"$

Combining the 2 steps calculations:
$M g^{2 +} ions = 3.00 mols M g C l_{2} \times \frac{1 mol M g^{2 +}}{1 mol M g C l_{2}} \times \frac{6.022 x 10^{23} M g^{2 +} i o n s}{1 mol M g^{2 +}} = 1.81 x 10^{24} M g^{2 +} ions$ $Mg^(2+)"ions"="3.00 mols " MgCl_2 xx ("1 mol " Mg^(2+))/("1 mol " MgCl_2)xx(6.022 " x " 10^23 Mg^(2+) ions)/("1 mol " Mg^(2+))=1.81" x "10^24 Mg^(2+) "ions"$

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How many $M g^{2 +}$ $Mg^(2+)$ ions are present in 3.00 moles of $M g C l_{2}$ $MgCl_2$ ?

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How many $M g^{2 +}$ $Mg^(2+)$ ions are present in 3.00 moles of $M g C l_{2}$ $MgCl_2$ ?