11. C. (Chong/this 1111.500.000.000.000.000.000.000.000.10.000.000. 0.02 mol of zine is added to 10.0cm^3 or 1.0moldm^-3 hydrochloric acid. Aln(s)+2NCl(aq)arrow V_(2)Cl_(2)(aq)+N_(2)(g) How many moles of hydrogen are produced? A 0.005 B.0.01 C.0.02 D. 0.04

To solve this problem, we need to use the balanced chemical equation provided:<br /><br />\[ \text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)} \]<br /><br />From the balanced equation, we can see that 1 mole of zinc (Zn) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of hydrogen gas (H₂).<br /><br />Given:<br />- 0.02 moles of zinc (Zn)<br />- The concentration of hydrochloric acid (HCl) is \(1.0 \, \text{mol/dm}^3\), which means there are \(1.0 \, \text{mol}\) of HCl in every \(1 \, \text{dm}^3\) (or \(1000 \, \text{cm}^3\)) of solution.<br /><br />First, we need to determine the volume of hydrochloric acid required to react with 0.02 moles of zinc. Since the reaction requires 2 moles of HCl for every mole of Zn, we need:<br /><br />\[ 0.02 \, \text{mol} \, \text{Zn} \times 2 \, \text{mol} \, \text{HCl/mol} \, \text{Zn} = 0.04 \, \text{mol} \, \text{HCl} \]<br /><br />Next, we calculate the volume of hydrochloric acid needed to provide 0.04 moles of HCl. Given the concentration of \(1.0 \, \text{mol/dm}^3\):<br /><br />\[ \text{Volume} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.04 \, \text{mol}}{1.0 \, \text{mol/dm}^3} = 0.04 \, \text{dm}^3 \]<br /><br />Since \(1 \, \text{dm}^3 = 1000 \, \text{cm}^3\), the volume in cm³ is:<br /><br />\[ 0.04 \, \text{dm}^3 \times 1000 \, \text{cm}^3/\text{dm}^3 = 40 \, \text{cm}^3 \]<br /><br />We are given that we have \(10.0 \, \text{cm}^3\) of hydrochloric acid. Since the available volume of hydrochloric acid (10.0 cm³) is less than the required volume (40 cm³), the limiting reagent is hydrochloric acid.<br /><br />Therefore, the amount of hydrogen gas produced is determined by the amount of hydrochloric acid available. Since we have \(10.0 \, \text{cm}^3\) of hydrochloric acid, we need to find out how much zinc is required to completely react with this amount of hydrochloric acid.<br /><br />From the balanced equation, 2 moles of HCl produce 1 mole of H₂. Therefore, 10.0 cm³ of 1.0 mol/dm³ HCl contains:<br /><br />\[ 10.0 \, \text{cm}^3 \times \frac{1.0 \, \text{mol}}{1000 \, \text{cm}^3} = 0.01 \, \text{mol} \, \text{HCl} \]<br /><br />Since 2 moles of HCl produce 1 mole of H₂, 0.01 moles of HCl will produce:<br /><br />\[ 0.01 \, \text{mol} \, \text{HCl} \times \frac{1 \, \text{mol} \, \text{H}_2}{2 \, \text{mol} \, \text{HCl}} = 0.005 \, \text{mol} \, \text{H}_2 \]<br /><br />Thus, the number of moles of hydrogen gas produced is:<br /><br />\[ \boxed{0.005} \]<br /><br />So, the correct answer is A. 0.005.