Answer:
CuSO4
Explanation:
Na2S + CuSO4 → Na2SO4 + CuS
The reaction is balanced (same number of elements in each side)
To determine limiting reagent you need to know the moles you have of each.
Molar mass Na2S = 23 * 2 + 32 = 78
Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5
Na2S mole = 15.5 / 78 = 0.2
CuSO4 mole = 12.1/159.5 = 0.076
*Remember mole = mass / MM
With that information now you have to divide each moles by its respective stoichiometric coefficient
Na2S stoichiometric coefficient : 1
Na2S : 0.2 / 1 = 0.2
CuSO4 stoichiometric coefficient: 1
CuSO4: 0.076 / 1 = 0.076
The smaller number between them its the limiting reagent, CuSO4
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
We can use combined gas laws to solve for the volume of the gas
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L
Matter - anything with mass and occupies space
accuracy - an indication of how close a measurement is to the correct result
precision - the degree to which a measurement can be replicated
meniscus - the curved top surface of a liquid column
volume - spaced occupied measured in cubic units
density - mass of an object per unit volume
Avagadro's number is just a measurement. One mole is 6.022 X 10^23 of anything - atoms, molecules, marbles... anything.
<span>1) If one mole = 6.022 X 10^23, then 8.00mol of H2S is: </span>
<span>(3.00mol H2S) (6.022 X 10^23 molecules H2S / 1 mol H2S) = 1.8060 X 10^24 molecules H2S. </span>
<span>Rounded to 3 sig figs =1.81 X 10^24 molecules H2S
</span>part2.
<span> This one uses moles in the stoichiometric sense as well as the measurement. One formula unit of MgCl2 contains 1 mole Mg and 2 moles Cl. </span>
<span>First, figure out how many moles of formula units there are. </span>
(1.81 X 10^24 FU's) (1mol MgCl2 / 6.022 X 10^23 FU's) = 3.0056mol MgCl2.
<span>Now, we know that there are 2 moles of Cl in every mole of MgCl2 (2 Cl atoms in every unit of MgCl2). From this we can determine how many moles of Cl atoms there are: </span>
<span>(3.0056mol MgCl2) (2mol Cl atoms / 1mol MgCl2) = 6.0112mol Cl atoms. </span>
<span>Now round to 3 sig figs = 10.0mol Cl atoms</span>
