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First you need to find out the Limiting reactant (LR). convert both reactants to the same thing. Check that the chemical equation is balanced. Now use stoichiometr and remember at moles, multiply: need moles, divide2 g / 42g/mol= 0.0477 mol propane mass propane/ Molar Mass propane = moles propane4 g / 32 g/mol= 0.125 mol oxygen X (1 mol/ 5 mol) = 0.025 mol propane oxygen is the LRmass O2 / MM O2 X (mol propane / mol O2)0.025 mol X (3 mol / 1 mol ) = .075 mol CO20.075 mol X (12 + 2*16) g /mol = 3.6 g CO2 In one step:2 g / 42g/mol X (3 mol / 1 mol ) X 48 g/mol = 6.86 g CO24 g / 32 g/mol X (3 mol / 5 mol ) X 48 g/mol = 3.6 g CO2mass/ MM X coefficient ratio X MM (new)
Answer:
<h3>D. </h3>
Explanation:
<h3>#I'm not sure</h3><h3>#happy learning</h3>
Heat can be absorbed or produced
The compound is Al2O3. The ratio of aluminum to oxygen is 2:3.