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2 HCl + 2 Mg ----> 2 MgCl + H2 this is the balanced equation of the reaction taking place.
to find the limiting reagent, convert the grams of HCl and Mg to moles of **THE SAME PRODUCT**. For instance, use the mole ratios and Molar masses (from the periodic table) to convert grams of HCl and Mg to moles H2. The original reactant that produces the least moles of the product is the limiting reagent, because all of it will be used up, or the actual amount produced. The excess reagent is the react that produced the larger moles of the product. I'll try to do the work below:
72 g HCl x (1mol HCl/36.46 g HCl) x (1 mol H2/2 mol HCl) = 0.987 mol H2
3.5 g Mg x (1 mol Mg/24.305 g Mg) x (1 mol H2/ 2 mol Mg) = 0.072 mol H2 *smaller mol, LR*
Knowing STP (273 K, 1 atm), use the ideal gas law (PV=nRT) to solve for the volume of hydrogen produced. In this case, use the R value of 0.0821 atmxL/molxK and the n (moles) value you found from the first part (make sure it is the mol H2 found from the limiting reagent!). Solve for volume, and you should get your answer in liters.
PV=nRT
(1 atm)(V)=(0.072 mol)(0.0821 atm x L/ mol x K)(273 K)
all the units, except for L, should cancel and leave you with:
0.620 L of H2 gas.
The calculations are hard to type, hopefully you can follow them fairly well. Check the numbers just incase, but this should be the right way to solve this problem. Good luck!
to find the limiting reagent, convert the grams of HCl and Mg to moles of **THE SAME PRODUCT**. For instance, use the mole ratios and Molar masses (from the periodic table) to convert grams of HCl and Mg to moles H2. The original reactant that produces the least moles of the product is the limiting reagent, because all of it will be used up, or the actual amount produced. The excess reagent is the react that produced the larger moles of the product. I'll try to do the work below:
72 g HCl x (1mol HCl/36.46 g HCl) x (1 mol H2/2 mol HCl) = 0.987 mol H2
3.5 g Mg x (1 mol Mg/24.305 g Mg) x (1 mol H2/ 2 mol Mg) = 0.072 mol H2 *smaller mol, LR*
Knowing STP (273 K, 1 atm), use the ideal gas law (PV=nRT) to solve for the volume of hydrogen produced. In this case, use the R value of 0.0821 atmxL/molxK and the n (moles) value you found from the first part (make sure it is the mol H2 found from the limiting reagent!). Solve for volume, and you should get your answer in liters.
PV=nRT
(1 atm)(V)=(0.072 mol)(0.0821 atm x L/ mol x K)(273 K)
all the units, except for L, should cancel and leave you with:
0.620 L of H2 gas.
The calculations are hard to type, hopefully you can follow them fairly well. Check the numbers just incase, but this should be the right way to solve this problem. Good luck!