Answers
Answer: 824.6 g of NaCl are produced from 500.0 g of chlorine.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex] [tex]\text{Moles of} Cl_2=\frac{500.00g}{71g/mol}=7.04moles[/tex]
[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]
According to stoichiometry :
1 mole of [tex]Cl_2[/tex] produce = 2 moles of [tex]NaCl[/tex]
Thus 7.04 moles of [tex]Cl_2[/tex] will produce=[tex]\frac{2}{1}\times 7.04=14.08moles[/tex] of [tex]NaCl[/tex]
Mass of [tex]NaCl=moles\times {\text {Molar mass}}=14.08moles\times 58.5g/mol=824.26g[/tex]
Thus 824.6 g of NaCl are produced from 500.0 g of chlorine.
Related Questions
a 5 g ice cube starts life at -1 C if 2182.5 j are added to it what will be the final temperature upon his demise ?
Answers
Answer:
Final Temp = 23.92°C
Explanation:
ΔH(total) = (m·c·ΔT)ice + (m·ΔH(f))melt'g + (m·c·ΔT)water
2182.5j = (5g)(2.092j/g·°C)(1°C) + (5g)(334.56j/g) + (5g)(4.184j/g·°C)(ΔT)
(2182.5 - 10.46 - 1672.8)j = 20.92j/°C·ΔT
ΔT = (2182.5 - 10.46 - 1672.8)j / 20.92j/°C = 23.92°C
Since the melting ice starts and ends at 0°C and is then warmed to 23.92°C then the temperature change is also the final temp of the water based upon given energy input values.
A reaction in the laboratory yields 5.98 g KAI(SO2)2, How many moles of potassium aluminum
sulfate were produced?
Answers
Answer:
0.0308 mol
Explanation:
In order to convert from grams of any given substance to moles, we need to use its molar mass:
Molar mass of KAI(SO₂)₂ = MM of K + MM of Al + (MM of S + 2*MM of O)*2Molar mass of KAI(SO₂)₂ = 194 g/molNow we calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g:
5.98 g ÷ 194 g/mol = 0.0308 mol