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QUESTION

# Using the reaction 2Na(s)+Cl2---&gt; 2NaCl(s), determine the mass of NaCl that is produced when a mixture of 12.8g Na and 10.2g Cl2 reacts? Determine the limiting reactant first

According to the reaction equation, 2 moles of Na requires 1 mole of Cl2 to produce 2 moles of NaCl. So we start by converting these moles to grams. Ma of Na is 23g/moles, and Ma for Cl2 is 71g/moles. 2 moles of Na equals to 2*23=46grams, and 1 mole Cl2 is 71 grams. Here we see that Na needs more Cl2 to react, so we can predict that 12.8 grams of Na will require more than 12.8 grams of Cl2, but we see that Cl2 is only 10.2 grams. This means that all of Cl2 will be consumed, some of Na will remain unreacted, which will make Cl2 the limiting reactant. For the second part, we first will calculate the mass of NaCl produced from 1 moles i.e. 71 grams of Cl2. Ma of NaCl is 58.5g/moles so 2 moles of NaCl will be 2*58.5=117grams. So 117 grams of NaCl is produced when 71 grams of Cl2 is consumed. Finally the NaCl produced when 10.2 grams of Cl2 is consumed is: 10.2*117/71=16.8grams.

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