10.0 grams of magnesium reacted with fluorine to produce MgF2
Find balance equation: 2MgF2=2MgF
1)can someone explain to me how many grams of fluorine reacted with magnesium.
2)How many grams of MgF2 were formed in the reaction?
To answer both questions, we need to use stoichiometry, which involves using the balanced chemical equation to relate the amounts of reactants and products.
1) To determine the number of grams of fluorine that reacted with magnesium, we need to first find the molar ratio between magnesium and fluorine in the balanced chemical equation. From the balanced equation, we can see that 1 mole of magnesium reacts with 1 mole of fluorine to produce 1 mole of MgF2.
Since we have 10.0 grams of magnesium, we need to convert this to moles. The molar mass of magnesium is approximately 24.31 g/mol, so:
10.0 g Mg * (1 mol Mg / 24.31 g Mg) = 0.411 mol Mg
Using the molar ratio from the balanced chemical equation, we can conclude that:
0.411 mol Mg * (1 mol F2 / 1 mol Mg) = 0.411 mol F2
To convert this to grams, we need to know the molar mass of fluorine, which is approximately 38.00 g/mol:
0.411 mol F2 * (38.00 g F2 / 1 mol F2) = 15.6 g F2
Therefore, approximately 15.6 grams of fluorine reacted with magnesium.
2) To find the number of grams of MgF2 formed in the reaction, we again consider the molar ratio between MgF2 and magnesium in the balanced equation. From the equation, we see that 1 mole of magnesium produces 1 mole of MgF2.
Using the calculated moles of magnesium (0.411 mol), we can determine the moles of MgF2 formed:
0.411 mol Mg * (1 mol MgF2 / 1 mol Mg) = 0.411 mol MgF2
To convert this to grams, we need to know the molar mass of MgF2, which is approximately 62.32 g/mol:
0.411 mol MgF2 * (62.32 g MgF2 / 1 mol MgF2) = 25.6 g MgF2
Therefore, approximately 25.6 grams of MgF2 were formed in the reaction.