The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>
The items that are true of early nuclear science are "the first nuclear reactions were done in the 1880s" and "the first nucleus split was uranium-235." <span>The answers are letters A and D. It is impossible that nucleus was lost during the reaction as it will not follow the law of mass conservation.
</span>
Answer:
Gravitational force is directly promotional to the mass
of both interacting objects.
Explanation:
More massive objects will attract each other with
a greater gravitional force. So as the mass of the
either objects increases the force of gravitational
attraction between them also increases.
False people can do whatever they want without people telling them what to do.
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
*
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
*
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be: