Question

Consider the following unbalanced equation for the reaction of aluminum with sulfuric acid.
Al(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2(g).

How many moles of H2 are formed by the complete reaction of 0.384 mol of Al?

Express your answer using three significant figures.

Chemical Equation -> 2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

Calculate the mass of water produced when 2.66 g of butane reacts with excess oxygen. Express your answer to three significant figures and include the appropriate units.

Calculate the mass of butane needed to produce 62.7 g of carbon dioxide. Express your answer to three significant figures and include the appropriate units.

Answer 1

2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)

From the balanced reaction ,

2 moles of Al upon reaction produces 3 moles of H2

0.384 moles of Al upon reaction produces M moles of H2

M = ( 0.384 x 3 ) / 2

= 0.576 moles of H2

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2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

Molar mass(g/mol) 58 32 18 44

2 mol = 2 x 58 g of butane produces 10 mol = 10 x 18 g of water

2.66 g of butane produces N g of water

N = ( 2.66x10x18) / (2x58)

= 4.13 g of water

8 mol = 8x44 g of CO2 produced from 2 mol = 2x58 g of butane

62.7 g of CO2 produced from ( 62.7 x 2x58)/(8x44) = 20.7 g of butane