please show work- These are the answers I got. Am I correct? What volume of oxyg

Question

please show work- These are the answers I got. Am I correct?

What volume of oxygen gas reacts with 5.00 mL of methane, CH4? (Assume temperature and pressure remain constant.) (2 Pts.)

        CH4(g) + 2 O2(g) (spark) CO2(g) + 2 H2O(g) (answer- 10ml)

Considering the limiting reactant, what is the mass of zinc sulfide (MM = 97.46 g/mol) produced from 0.750 g of zinc and 0.250 g of sulfur? (3 Pts.)

     Zn(s) + S(s) (spark) ZnS(s) (Answer:0.0026)

Considering the limiting reactant, what is the volume of NO gas produced from 50.0 L of ammonia gas and 60.0 L of oxygen gas? (Assume constant conditions.)

4NH3(g) + 5O2(g)    4NO(g) + 6H2O(g) (4 Pts.) (Answer: 48L)

What is the mass of calcium phosphate, Ca3(PO4)2, that can be prepared from 2.50 g of Na3PO4? (4 Pts.)

2Na3PO4 (aq) + 3Ca(OH)2 (aq) Ca3(PO4)2 (s) + 6NaOH (aq) (Answer: 2.356g)

Explanation / Answer

CH4(g) + 2 O2(g) (spark) CO2(g) + 2 H2O(g)

1 mole of CH4 react with 2 moles of O2

22400ml of Ch4 react with 2*22400ml of O2

5 ml of CH4 react with = 2*22400*5/22400 = 10ml of O2

Zn(s) + S(s) (spark) ZnS(s)

no of moles of Zn = W/G.A.Wt = 0.75/65.4   = 0.0115 moles

no of moles of S    = W/G.A.Wt     = 0.25/32    = 0.0078moles

1 mole of Zn react with 1 mole of S

S is limiting reagent

1 mole of S react with Zn to gives 1 mole of ZnS

0.0078 moles of S react with Zn to gives 0.0078 moles of ZnS

mass of ZnS = no of moles * gram molar mass

                      = 0.0078*97.46 = 0.76g >>>>answer

4NH3(g) + 5O2(g)    4NO(g) + 6H2O(g)

4 moles of NH3 react with 5 moles of O2

4*22.4 L of NH3 react with 5*22.4 L of O2

50L of NH3 react with = 5*22.4*50/4*22.4   = 62.5L of O2

O2 is limiting reagent

5 moles of O2 react with NH3 to gives 4 moles of NO

5*22.4L of O2 react with NH3 to gives 4*22.4L of NO

60L of O2 react with NH3 to gives = 4*22.4*60/5*22.4   = 48L of N

2Na3PO4 (aq) + 3Ca(OH)2 (aq) Ca3(PO4)2 (s) + 6NaOH

2 moles of Na3PO4 react with Ca(OH)2 to gives 1 mole of Ca3(PO4)2

2*164g of Na3PO4 react with Ca(OH)2 to gives 310g of Ca3(PO4)2

2.5g of Na3PO4 react with Ca(OH)2 to gives = 310*2.5/2*164   = 2.36g of Ca3(PO4)2

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