How would you balance Na(s) + Cl2 → NaCl(s)? - Socratic
2015年10月18日 · 1 Answer. The balanced equation is 2Na (s)+Cl2(g) → 2NaCl (s). How to Balance the Equation Na (s)+Cl2(g) → NaCl. Since there are two chlorine atoms on the left side, there needs to be two chlorine atoms on the right side. So I put a coefficient of two in front of the formula NaCl. Na (s)+Cl2(g) → 2NaCl.
Na + Cl2 --> 2 NaCl If 15 mol Na reacts with 15 mol Cl2 , _ is the ...
2018年2月25日 · You forgot to balance the equation for each mol of Cl_2, 2 mols of Na react, so therefore Na is limiting and Cl_2 is excess Check for LR 15 mol Cl_2 *(2mol Na)/ (1 mol Cl_2) Therefore 30 mol Na will be required to react, and since we don't have that amount, Na is limiting.
In the reaction 2Na + Cl2 --> 2NaCl Would 2Na be the ... - Socratic
2018年4月14日 · Just Na. We have the balanced equation: 2Na(s)+Cl_2(g)->2NaCl(s) Na(0) gets oxidized to Na(I), while Cl(0) gets reduced to Cl(-I). Since sodium got oxidized, it is the reducing agent in this equation. It doesn't matter if there was one sodium or three sodium atoms, the reducing agent is just the element or molecule itself, so Na will be the reducing agent here.
What is the equation for sodium and chlorine react to form …
2017年3月14日 · 2Na + Cl_2 -> 2NaCl The skeletal equation: Na + Cl_2 -> NaCl Balanced: 2Na + Cl_2 -> 2NaCl
What are the grams of sodium chloride produced when 10.0 g
2017年8月16日 · m (NaCl)=16.479g 1) We need to know which of reagents is excess, so we calculate it by reagents' mass. M (Na)=23 frac\\{g}{mol} m (Na)=n×M=2×23=46 g M (Cl_2)=35.5×2=71 frac\\{g}{mol} m (Cl_2)=n×M=1×71=71 g \\frac{10}{46}:\\frac{x}{71} x=\\frac {10×71}{46}=15.43 This means the sodium is excess. So, we're continuing solving the problem by chlorine. 2) M (NaCl)=23+35.5=58.5 frac\\{g}{mol} m ...
How many moles of chlorine gas (Cl2) would react with 5
2015年11月4日 · How many grams of NaOH is produced from #1.20 x 10^2# grams of #Na_2O#? How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH? Na2O + H2O ---> 2 NaOH
Sodium and chlorine gas react to produce sodium chloride. If
2017年5月20日 · "92.1 g" You know that 2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s)) and that the reaction produced "234 g" of sodium chloride. Convert this to moles by using the compound's molar mass 234 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44 color(red)(cancel(color(black)("g")))) = "4.004 moles NaCl" Now, you know that the sample of sodium metal reacted completely, which implies that ...
Sodium and chlorine react to form sodium chloride: 2Na (s) + Cl2 …
2017年6月15日 · The limiting reactant is "Na". The theoretical yield is 112 g "NaCl". > This is a limiting reactant problem. We know that we will need a balanced equation with masses, molar masses, and moles of the compounds involv
Stoichiometry - Chemistry - Socratic
Stoichiometry is the chemistry that mathematically relates all substances in a reaction, quantitatively relating the amount of reactants and products in a chemical reaction. It allows the chemist to determine the amount of product that will form from a given amount of reactants, or the amount of one reactant that is needed to react completely with some specific amount of …
For the reaction 2NaI + Cl2 →2NaCl + I2 How many grams of NaCl …
2018年3月12日 · This will make NaCl: 0.0333 moles #NaI xx ("2 mol NaCl"/"2 mol NaI")# NaCl moles = 0.0333 moles . With the #Cl_2#, we can make 0.1410 moles NaCl. Using the NaI, we can only make 0.0333 moles of NaCl. This means we will run out of NaI....NaI is our limiting reagent, and it tells us that using 5g Cl2 and 5g of NaI, we can only make 0.0333 moles NaCl.