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How many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous oxide? HNO3 + Cu2O ---> Cu(NO3)2 + NO + H2O (unbalanced)
Answers
Answer:
] How many grams of calcium phosphate can be produced from the reaction of 2.50 L of 0.250 M Calcium chloride
with and excess of phosphoric acid?
Calcium chloride + phosphoric acid --> calcium phosphate + hydrochloric acid
3CaCl2 + 2H3PO4 ---> Ca3(PO4)2 + 6HCl
2.50L CaCl2 X.
0.250L CaCl2
1mol X. 3mol CaCl2
1mol Ca3(PO4)2 X. 1mol
310.0g Ca3(PO4)2 =
Answer ____________________ 64.6g Ca3(PO4)2
2] How many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous oxide
14 HNO3 + 3Cu2O ---> 6 Cu(NO3)2 + 2 NO + 7H2O
100.0g Cu2O X
1mol Cu2O
143.1g X
14mol HNO3
3mol Cu2O X
1000ml HNO3
1.50mol HNO3
= 2.18 x 103ml HNO3
Answer _________________________
3] 60.5 mL of HNO3 are required to react with 25.0 mL of a 1.00 M Barium hydroxide solution:
2 HNO3(aq) + Ba(OH)2(aq) --> 2 H2O(s) + Ba(NO3)2(aq) (BALANCED)
Find the Molarity of the nitric acid solution
25.0ml Ba(OH)2 X 1L
103ml X. 1L
1.00mol Ba(OH)2 X
2mol HNO3
1mol Ba(OH)2
= 0.0500mol HNO3
MHNO3 =
0.0500 mol HNO3
0.0605 L soln = 0.826 M
Answer _________________________
4] For the following equation determine which reactant is the limiting reactant and which reactant is in excess. The
amounts of reagent used are shown. Show calculations to support your choices
3Fe + 4H2O ----> Fe3O4 + 4H2
40.0 g 16.0g
40.0g Fe X 1molFe
55.8g X
1mol Fe3O4
3molFe = 0.239 mol Fe3O4
16.0g H2O X
1molH2O
18.0g X
1mol Fe3O4
4mol H2O = 0.222 mol Fe3O4 <==== amount made
The limiting reactant is ____H2O_____ The excess reactant is ______Fe______
5] 35.5 g of silver nitrite is reacted with 35.5 grams of sodium sulfide which produces silver sulfide and sodium nitrite.
a. Write and balance the equation 2AgNO2 + Na2S ---> Ag2S + 2NaNO2
b.. Calculate the number of grams of silver sulfide produced.
35.5g AgNO2 X 1mol AgNO2
153.9g X 1mol Ag2S
2mol AgNO2 = 0.115 mol Ag2S<==== amount made
35.5g Na2S X.
1mol Na2S
78.0g X 1mol Ag2S
1mol Na2S = 0.455 mol Ag2S
0.115 mol mol Ag2S X 247.8g Ag2S
1mol = 28.5g Ag2S
Answer:
How many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous