What mass of chromium (cr) is contained in 35.8 g of (nh4)2cr2o7 ?

M{(NH₄)₂Cr₂O₇}=2A(N)+8A(H)+2A(Cr)+7A(O)

M{(NH₄)₂Cr₂O₇}=252.065 g/mol

M(Cr)=51.996 g/mol

m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}=m(Cr)/2M(Cr)

m(Cr)=2M(Cr)m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}

m(Cr)=2*51.996*35.8/252.065=14.770 g

m(Cr)=14.770 g

The remaining mass of ethane is 3.19 grams

Explanation:

Step 1: Data given

Mass of ethane = 7.22 grams

Mass of oxygen = 15.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = 7.22 grams / 30.07 g/mol

Moles ethane = 0.240 moles

Step 4: Calculate moles oxygen

Moles oxygen = 15.0 grams / 32.0 g/mol

Moles oxygen = 0.46875 moles

Step 5: Calculate limiting reactant

For 2 moles ethane we need 7 moles oxygen to produce 4 moles CO2 and 6 moles H2O

Oxygen is the limiting reactant. It will completely be consumed (0.46875 moles). Ethane is in excess. There will react 0.46875 / 3.5 = 0.134 moles

There will remain 0.240 - 0.134 = 0.106 moles

Step 6: Calculate mass of remaining ethane

Mass ethane = 0.106 moles * 30.06 g/mol

Mass ethane = 3.19 grams

The remaining mass of ethane is 3.19 grams