It’s lower, hope this helps with your issue and helps solve your problem, no problem
Ice is less dese than water
Elements that belong to the same period which is seen horizontally in the periodic table have different properties. Elements with similar properties are seen in columns or groups. Elements in the same period have similar principal quantum number and belong to the same principal shell. Hence the answer is that t<span>hey have different properties that repeat across the next period</span>.
From the coefficients of the equation, we know that for every 3 moles of water consumed, 1 mole of diphosphorus trioxide is consumed.
This means we need to find the mass of 0.75 moles of diphosphorus trioxide.
- The atomic mass of phosphorous is 30.973761998 g/mol.
- The atomic mass of oxygen is 15.9994 g/mol.
So, the formula mass of diphosphorus trioxide is:
- 2(30.973761998)+3(15.9994)=109.945723996 g/mol.
Thus, 0.75 moles have a mass of:
- 0.75(109.945723996), which is about 82.5 g (to 3 sf)
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: