Question

magnesium (24.30 g) reacts with hydrogen chloride (x g) to produce hydrogen gas (2.04 g) and magnesium chloride (96.90 g). what mass of hydrogen chloride was used in the reaction? round your answer to the second decimal.

Explanation:

Step1: Write the chemical reaction equation

$Mg + 2HCl=MgCl_{2}+H_{2}$

Step2: Calculate the moles of magnesium

The molar - mass of $Mg$ is $M_{Mg}=24.30\ g/mol$. The moles of $Mg$, $n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{24.30\ g}{24.30\ g/mol}=1\ mol$.

Step3: Calculate the moles of hydrogen gas

The molar - mass of $H_{2}$ is $M_{H_{2}} = 2.04\ g/mol$. The moles of $H_{2}$, $n_{H_{2}}=\frac{m_{H_{2}}}{M_{H_{2}}}=\frac{2.04\ g}{2.04\ g/mol}=1\ mol$.

Step4: Calculate the moles of magnesium chloride

The molar - mass of $MgCl_{2}$ is $M_{MgCl_{2}}=96.90\ g/mol$. The moles of $MgCl_{2}$, $n_{MgCl_{2}}=\frac{m_{MgCl_{2}}}{M_{MgCl_{2}}}=\frac{96.90\ g}{96.90\ g/mol}=1\ mol$.

Step5: Use the mole - ratio from the balanced equation

From the balanced equation $Mg + 2HCl=MgCl_{2}+H_{2}$, the mole - ratio of $HCl$ to $Mg$ is $2:1$. Since $n_{Mg} = 1\ mol$, the moles of $HCl$ used, $n_{HCl}=2n_{Mg}=2\ mol$.

Step6: Calculate the mass of $HCl$

The molar - mass of $HCl$ is $M_{HCl}=1.01\ g/mol + 35.45\ g/mol=36.46\ g/mol$. The mass of $HCl$, $m_{HCl}=n_{HCl}\times M_{HCl}=2\ mol\times36.46\ g/mol = 72.92\ g$.

Answer:

$72.92$