$M g$ $Mg$ and $O_{2}$ $O_2$ react in a 2.1 molar ratio. 2 moles $M_{g}$ $M_g$ = 1 mole $O_{2}$ $O_2$ . If a reaction used 32.5 g of $O_{2}$ $O_2$ , how many g of $M g$ $Mg$ reacted?

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Answer 1 · 2016-03-16T02:31:59

$49.4 g of Mg$ $"49.4 g of Mg"$ reacted.

Step 1. Calculate the moles of $O_{2}$ $"O"_2$ .

$32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2$

Step 2. Use the molar ratio to calculate the moles of $"Mg"$ .

$1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"$

Step 3. Calculate the mass of $"Mg"$ .

$2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"$

The reaction required $"49.4 g Mg"$ .

MgMgMg and O_2O2O_2 react in a 2.1 molar ratio. 2 moles M_gMgM_g = 1 mole O_2O2O_2. If a reaction used 32.5 g of O_2O2O_2, how many g of MgMgMg reacted?