Stoichiometry

how many moles of potassium oxide (K2O) will be formed when 1.34 moles of potassium reacts with oxygen according to 4K+O2 --> 2K2O

• 0.67 moles K2O

• 6.7 moles K2O

• 67 moles K2O

• 0.34 moles K2O

Tags

NGSS.HS-PS1-7

How many moles of copper are needed to react with sulfur to produce 1.34 moles of copper (I) sulfide according to the following reaction 2Cu + S --> 2 K2O

• 26.8 moles Cu

• 2.68 moles Cu

• 268. moles Cu

• 4.67 moles Cu

How many moles of CO2 are produced from the combustion of 4.60 mol C3H8

1 C3H8 + 5O2 --> 2 CO2 + 4 H2O

• 13.8 mol CO2

• 1.38 mol CO2

• 138. mol CO2

• 18.3 mol CO2

How many moles of CS2 forms when 5.6 mol C reacts

5C + 2SO2 --> CS2 + 4CO

• 1.12 mol CS2

• 11.2 mol CS2

• 112. mol CS2

• 3.16 mol CS2

Tags

NGSS.HS-PS1-7

How many grams of iron (III) oxide are produced when 257.4 grams of FeS are used?

4FeS + 7O2 --> 2Fe2O3 + 4SO2

• 2.336 g iron (III) oxide

• 23.36 g iron (III) oxide

• 233.6 g iron (III) oxide

• 156.8 g iron (III) oxide

Tags

NGSS.HS-PS1-7

How many grams of NO is produced if 26.0g of O2 is combined with excess ammonia?

4NH3 + 5O2 --> 6H2O + 4NO

• 17.8g NO

• 195g NO

• 1.95g NO

• 19.5g NO

Tags

NGSS.HS-PS1-7

If 75.2g of Copper reacts with 39.8g of Silver Nitrate, according to the following reaction, what is the maximum number of grams of silver that can be produced?

1CU + 2Ag(NO3) --> 1CU(NO3)2 + 2 Ag

• 39.8 g Ag(NO3)

• 3.98 g Ag(NO3)

• 398 g Ag(NO3)

• 45.7 g Ag(NO3)

Tags

NGSS.HS-PS1-7

if 145.0g of NH4Cl reacts with 68g of Ca(OH)2 according to the following reaction, what is the maximum mass of ammonia that can be formed?

2NH4Cl + Ca(OH)2 --> CaCl2 + 2NH3 + 2H2O

• 311.8g of ammonia

• 31.18g of ammonia

• 3118g of ammonia

• 28.61g of ammonia

Tags

NGSS.HS-PS1-7

if 75.0g of magnesium is reacted with an excess of nitric acid, then 2.9g of hydrogen is actually produced, what is the percent yield of hydrogen

1Mg + 2 HNO3 --> 1Mg(NO3)2 + 1H2

• 43.0%

• 4.70%

• 47.0%

• 23.0%

Tags

NGSS.HS-PS1-7

if 17.0g aluminum was reacted with an excess of bromine to yield 100. aluminum bromide find the theoretical yield then calculate the percent yield

2Al(s) + 3Br2(L) --> AlBr3(s)

• 535%

• 47.9%

• 53.5%

• 59.5%

Tags

NGSS.HS-PS1-7