The chemical reaction is:
2 Mg + O2 = 2 MgO
Mg+dil2HCl----> MgCl2 + H2
First you must find out what mass of each would react perfectly, then, if you have more than is needed of one of the reactants (if it is in excess) all of the reactant will react. Here is the calculation you need, for example, say you have 50g of each reactant. Step 1-Write out formula of reactants Mg + O2 = MgO2 1mole 1 mole Step 2 - Find the gram formula mass of reactants 1 mole Mg= 24.3 g 1 mole O2= 32 g 24.3g Mg reacts with 32g O2 Step 3 - Find amount required to react 50g Mg -- 50/32x24.3 =37.9 =37.9g Mg From that we can see that since there is 37.5g Mg and only 24.3g is needed to react completely with O2, the Mg is in excess. Substitute your starting weights in there and use that calculation, and add more than the required amount, that way you can be sure.
Do you mean this reaction? 2Mg + O2 -> 2MgO 4.11 moles Mg (1 mole O2/2 mole Mg) = 2.06 moles oxygen gas consumed --------------------------------------------------
The chemical equation is:2 Mg + O2 = 2 MgO
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2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
First you must find out what mass of each would react perfectly, then, if you have more than is needed of one of the reactants (if it is in excess) all of the reactant will react. Here is the calculation you need, for example, say you have 50g of each reactant. Step 1-Write out formula of reactants Mg + O2 = MgO2 1mole 1 mole Step 2 - Find the gram formula mass of reactants 1 mole Mg= 24.3 g 1 mole O2= 32 g 24.3g Mg reacts with 32g O2 Step 3 - Find amount required to react 50g Mg -- 50/32x24.3 =37.9 =37.9g Mg From that we can see that since there is 37.5g Mg and only 24.3g is needed to react completely with O2, the Mg is in excess. Substitute your starting weights in there and use that calculation, and add more than the required amount, that way you can be sure.
Do you mean this reaction? 2Mg + O2 -> 2MgO 4.11 moles Mg (1 mole O2/2 mole Mg) = 2.06 moles oxygen gas consumed --------------------------------------------------
The chemical equation is:2 Mg + O2 = 2 MgO
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The chemical reaction equation would be Mg + CO2 + O -> ?.
The reaction is between Mg and O2. 2Mg+O2->2MgO
2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
CH4 + 2O2 gives CO2 and 2H2O.So 16g of CH4 react with 64g of O2.So 24g react with 96g of O2
how do metals react with oxygen
Mg+O2----->MgO
I'm not entirely sure but an unbalanced equation is: Mg + O2 -----> MgO
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)