Solving Chemistry Homework: Calcium Chloride, KMnO4, NaOH, HCl, KCIO3, NH4PO4

[alexx4]

Homework Statement

4.How many grams of calcium chloride can be prepared if a student uses 500g if calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution in the reaction.

5. 2KMnoO4 + 5 H2C2O4<----need to use this

Concentration of 0.24L KMnO4 solution required to oxidize 2.7g of Oxalic Acid?

6.What volume of 0.100mol/L aqueous sodium hydroxide(NaOH) is neutralized by 500ml of 0.200mol/L hydrochloric acid(HCI)

7.When 10.0g of potassium chlorate,KCIO3,decomposes,what volume of oxygen is obtained(at 18C and 750 mm HG pressure)

8. 2(NH4)3PO4-->3N2+O2+12H2+2PO3

a)name products
b)Name the reactant
c)type of reaction

Homework Equations

The Attempt at a Solution

 

[Nino MMP]

4. CaCO3 + 2NaCl --> CaCl2 + Na2CO3 moles of CaCO3= 500/100.09 = 4.988 moles of NaCl = 2/1000 * 4 = 8 moles of CaCl2 = 4.988 Mass of CaCl2 = 4.988* 110.98 = 550.5g of CaCl2 5. Moles of Oxalic acid = 2.7/90.03 = 0.03 Volume = 0.024/0.03 = 0.8L of KMnO4 solution 6.Moles of HCI = 500/1000 * 0.200 = 0.1 Moles of NaOH = 0.1 Volume of NaOH = 0.1/0.100 = 1L 7.moles of KCIO3 = 10/122.55 = 0.08166 Volume of oxygen = 0.08166 * 22.4 = 1.86L 8.a)Nitrogen, Oxygen, Hydrogen and Phosphorus b)Ammonium phosphate c)Decomposition

 

[Blueshift5]

4. To solve this problem, we first need to write out the balanced chemical equation for the reaction between calcium carbonate and sodium chloride:

CaCO3 + 2NaCl --> CaCl2 + Na2CO3

Next, we need to calculate the moles of sodium chloride present in the 2.00 L solution:

2.00 L x 4.00 mol/L = 8.00 mol NaCl

Using the mole ratio from the balanced equation, we can determine the moles of calcium carbonate:

8.00 mol NaCl x (1 mol CaCO3 / 2 mol NaCl) = 4.00 mol CaCO3

Finally, we can convert the moles of calcium carbonate to grams using its molar mass:

4.00 mol CaCO3 x 100.09 g/mol = 400.36 g CaCO3

Therefore, the student can prepare 400.36 grams of calcium chloride from 500 g of calcium carbonate and 2.00 L of a 4.00 Mol/L sodium chloride solution.

5. To solve this problem, we first need to write out the balanced chemical equation for the reaction between potassium permanganate and oxalic acid:

2KMnO4 + 5H2C2O4 --> 2MnC2O4 + 2K2C2O4 + 3CO2 + 5H2O

Next, we need to determine the moles of oxalic acid present in 2.7 g:

2.7 g x (1 mol / 90.03 g) = 0.03 mol H2C2O4

Using the mole ratio from the balanced equation, we can determine the moles of potassium permanganate:

0.03 mol H2C2O4 x (2 mol KMnO4 / 5 mol H2C2O4) = 0.012 mol KMnO4

Finally, we can use the concentration and volume of the KMnO4 solution to calculate the required volume:

0.012 mol / 0.24 mol/L = 0.05 L or 50 mL

Therefore, 50 mL of the 0.24 L KMnO4 solution is required to oxidize 2.7 g of oxalic acid.

6. To solve this problem, we first need to write out the balanced chemical equation for the neutralization