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Question
question 8 (1 point) hydrogen and oxygen react chemically to form water. how much water would form if 14 kg of hydrogen reacted with 34 kg of oxygen? h₂ + o₂ → h₂o do not include the units in your answer ______ question 9 (1 point) when ammonium nitrate nh₄no₃ explodes, the products are nitrogen, oxygen, and water. when 40 grams of ammonium nitrate explodes, 14 grams of nitrogen and 8 grams of oxygen form. how many grams of water form? nh₄no₃ → n₂ + o₂ + h₂o ______
Step1: Balance the reaction equation
The balanced equation for hydrogen and oxygen reacting to form water is $2H_2 + O_2
ightarrow 2H_2O$.
Step2: Calculate the moles of reactants
The molar mass of $H_2$ is approximately $2\ g/mol$, so for $14\ g$ of $H_2$, the number of moles $n_{H_2}=\frac{14\ g}{2\ g/mol}=7\ mol$. The molar mass of $O_2$ is approximately $32\ g/mol$, so for $34\ g$ of $O_2$, the number of moles $n_{O_2}=\frac{34\ g}{32\ g/mol}\approx1.0625\ mol$.
Step3: Determine the limiting reactant
From the balanced - equation, the mole ratio of $H_2$ to $O_2$ is $2:1$. For $7\ mol$ of $H_2$, we would need $n_{O_2\ required}=\frac{7\ mol}{2}=3.5\ mol$ of $O_2$. Since we have only $1.0625\ mol$ of $O_2$, $O_2$ is the limiting reactant.
Step4: Calculate the moles of water formed
The mole ratio of $O_2$ to $H_2O$ is $1:2$. So the number of moles of $H_2O$ formed $n_{H_2O}=2\times n_{O_2}=2\times1.0625\ mol = 2.125\ mol$.
Step5: Calculate the mass of water formed
The molar mass of $H_2O$ is approximately $18\ g/mol$. So the mass of $H_2O$ formed $m_{H_2O}=n_{H_2O}\times18\ g/mol=2.125\times18 = 38.25\ g$.
For the second part:
Step1: Apply the law of conservation of mass
According to the law of conservation of mass, the mass of reactants equals the mass of products in a chemical reaction. The mass of ammonium nitrate is $40\ g$, and the masses of nitrogen and oxygen formed are $14\ g$ and $8\ g$ respectively.
Let the mass of water formed be $m$. Then $40\ g=14\ g + 8\ g+m$.
Step2: Solve for the mass of water
$m=40-(14 + 8)=18\ g$.
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38.25
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