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What is the reaction $2Mg(s) + O_2(g) -> 2MgO(s)$ is an example of what type of reaction?
Magnesium metal is oxidized to magnesium ion: $Mg(s) rarr Mg^(2+) + 2e^(-)$ $(i)$ Oxygen gas is reduced to oxide anion: $O_2(g) + 4e^(-) rarr 2O^(2-)$ $(ii)$ Overall $2xx(i) + (ii)$,...
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In the reaction $H_2S + 2O_2 -> H_2SO_4$, the law of definite proportions predicts that for every mole of $H_2S$, you will need how many moles of $O_2$?
I would look at your equation again. You have even balanced and formatted it properly. What does it say? It says 1 molecule of hydrogen sulfide reacts with 2 molecules...
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How many atoms of oxygen are there on each side of the chemical equation $2Mg + O_2 -> 2MgO$?
The expression $2Mg + O_2 -> 2MgO$ Means Two atoms of magnesium, plus one oxygen molecule (consisting of two atoms of oxygen) react to become two particles of magnesium oxide,...
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17.2 grams $MgO$ are produced according to the following equation: $2Mg + O_2 -> 2MgO$. How many grams of oxygen $O_2$ must have been consumed?
Balanced Equation $"2Mg(s)" + "O"_2("g")$$rarr$$"2MgO(s)"$ You need the molar masses of $"MgO"$ and $"O"_2"$. Molar Masses $"MgO":$$"40.304400 g/mol"$ )%22 $"O"_2:$$"31.9988 g/mol"$ Determine the moles of MgO by dividing the...
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$Mg$ and $O_2$ react in a 2.1 molar ratio. 2 moles $M_g$ = 1 mole $O_2$. If a reaction used 32.5 g of $O_2$, how many g of $Mg$ reacted?
Step 1. Calculate the moles of $"O"_2$. $32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2$ Step 2. Use the molar ratio to calculate the...
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How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen in the reaction $2Mg + O_2 -> 2MgO$?
Let us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium ,...
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Magnesium will burn in oxygen to form magnesium oxide as represented by the following equation
2Mg(s) + O2(g) -> 2MgO(s)
What mass of magnesium will react with 498.1 ml of oxygen at 145.2 degrees Celsius and 0.927 atm?
We use the Ideal Gas Equation to access the molar quantity of dioxygen gas............ $n=(PV)/(RT)=(0.927*atmxx0.4981*L)/(0.0821*(L*atm)/(K*mol)xx418.4*K)=1.34xx10^-2*mol$. Given the stoichiometric equation.......... $Mg(s) + O_2(g) rarr MgO(s)$ There are $1.34xx10^-2*molxx24.31*g*mol^-1=??*g$ metal required for...
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Consider the following reaction: $2Mg(s) + O_2(g) -> 2MgO(s)$, $DeltaH = -1204kJ$. Is this reaction exothermic or endothermic? How do you calculate the amount of heat transferred when 2.4 grams of Mg(s) reacts at constant pressure?
We has.............. $2Mg(s) + O_2(g) rarr 2MgO(s)$ $DeltaH^@""_"rxn"=-1204*kJ*mol^-1$ $(i)$ To make the arithmetic a bit easier we could write........ $Mg(s) + 1/2O_2(g) rarr MgO(s)$ $DeltaH^@""_"rxn"=-602*kJ*mol^-1$ $(ii)$ I am CLEARLY free...
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Given....$2Mg(s) + O_2(g) rarr 2MgO(s)$ $; DeltaH_"rxn"=-1203*kJ*mol^-1$
What enthalpy is associated with the combustion of FOUR moles of magnesium metal?
$2Mg(s) + O_2(g) rarr 2MgO(s)$ $ DeltaH_"rxn"=-1203*kJ*mol^-1$ Enthalpy terms are written per mole of reaction as written. We combust 4 moles of magnesium metal, and should therefore generate $2xx-1203*kJ=-2406*kJ$. Note...
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What is the reduction half-reaction for $2Mg + O_2 -> 2MgO$?
Start by assigning to all the atoms that take part in the reaction--it's actually a good idea to start with the unbalanced chemical equation $stackrel(color(blue)(0))("Mg")_ ((s)) + stackrel(color(blue)(0))("O") _...
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