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Water can be formed from the stoichiometric reaction of hydrogen with oxygen: $2H_2(g) + O_2(g) + 2H_2O(g)$. A complete reaction of 5.0 g of $O_2$ with excess hydrogen produces how many grams of $H_2O$?
$2H_2(g) + O_2(g) rarr 2H_2O(g)$ We started with $(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"$. Given excess dihydrogen, this molar quantity thus gives $2xx0.156*mol$ water. And thus, $2xx0.156*molxx18.01*g*mol^-1~=6*g$ water are evolved. Is this reaction exothermic...
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The reaction between $H_2$ and $O_2$ produces 13.1 g of water. How many grams of $O_2$ reacted? $2H_2(g) + O_2(g) -> 2H_2O(g)$?
There are $(13.1*g)/(18.01*g*mol^-1)=0.727*mol$ of water product. $H_2(g) + 1/2O_2(g) rarr H_2O(l)$ Given the of the reaction, there were necessarily $0.727*mol$ dihydrogen reactant, and $(0.727*mol)/2$ dioxygen reactant. I speak of dihydrogen,...
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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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How many molecules of $O_2$ are required for the complete combustion of 23.0 grams of hexane in the reaction $C_6H_14 + O_2 -> CO_2 + H_2O$?
The complete combustion requires $1.527× 10^24color(white)(l) "molecules O"_2$. There are four steps to follow in this problem: Balance the equation. Use the molar mass of hexane to calculate...
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How many moles of $CO$ and $O_2$ are required to produce 10 moles of $CO_2$ in the reaction $2CO + O_2 -> 2CO_2$?
Balanced equation $"2CO"("g") + "O"_2("g")"$$rarr$$"2CO"_2("g")"$ To calculate moles $"CO"$ and $"O"_2"$ required to produce $"10 mol CO"_2"$, multiply mol $"CO"_2"$ by the mol ratio between the desired reactant and $"CO"_2"$...
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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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Given the equation $2H_2O ->2H_2 + O_2$, how many moles of $H_2O$ would be required to produce 2.5 moles of $O_2$?
Since the chemical equation, $color(red)2H_2O->2H_2+color(blue)1O_2$ requires a specific number of moles of the reactant, then a specific number of moles of the products are created. With this knowledge,...
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Given the equation $2H_2O -> 2H_2 + O_2$, how many moles of $H_2O$ be required to produce 2.5 moles of $O_2$?
So every 2 molecules of water gives 1 molecule of oxygen. This ratio ($2:1$) stays the same if we are talking about moles, because a mole is a set number...
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$2H_2O -> 2H_2 + O_2$ What is the percent yield of $O_2$ if 10.2 g of $O_2$ is produced from the decomposition of 17.0 g of $H_2O$?
You know by looking at the balanced chemical equation $2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))$ that it takes $2$ moles of water to produce...
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A sample containing 4.80 g of $O_2$ gas has a volume of 15.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.500 mole of $O_2$ gas is added to the initial sample of 4.80 g of $O_2$?
Initially, you have $4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2$ Thus, you have $"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"$ If you...
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